Download Study Guide Advanced Algebra Semester Final 12/16/2009 Direct

Study Guide
Advanced Algebra Semester Final
12/16/2009
Direct Variation
Variation equations are formulas that show how one quantity changes in relation to one or more other
quantities. There are four types of variation: direct, indirect (or inverse), joint, and combined.
Direct variation equations show a relationship between two quantities such that when one quantity increases, the
other also increases, and when one quantity decreases, the other also decreases. We can say that y varies
directly as x, or y is proportional to x. Direct variation formulas are of the form y = kx, where the number
represented by k does not change and is called a constant of variation.
Indirect variation equations are of the form y = k/x and show a relationship between two quantities such that
when one quantity increases, the other decreases, and vice versa.
This skill focuses on direct variation. The following is an example of a direct variation problem.
The amount of money in a paycheck, P, varies directly as the number of hours, h, that are worked. In
this case, the constant k is the hourly wage, and the formula is written P = kh. If the equation is solved
for k, the resulting equation shows that P and h are proportional to each other.
Therefore, when two variables show a direct variation relationship, they are proportional to
each other. Direct variation problems can be solved by setting up a proportion in the form below.
Example 1: The amount of fuel needed to run a textile machine varies directly as the number of hours
the machine is running. If the machine required 8 gallons of fuel to run for 24 hours, how many gallons
of fuel were needed to run the machine for 72 hours? Round your answer to the nearest tenth of a
gallon, if necessary.
Step 1: Set up the proportion. Since the machine used 8 gallons of fuel in 24 hours, the left side of the
proportion should be 8 gallons over 24 hours. The number of gallons that the machine used in 72 hours
needs to be found, so the right side of the proportion should be g gallons over 72 hours.
Step 2: Cross-multiply across the equal sign.
Step 3: Set up the cross-multiplication equation.
Step 4: Divide both sides of the equation by 24 hours to isolate g gallons.
Step 5: Reduce the fractions on both sides of the equal sign.
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Step 6: Simplify by multiplying the numbers remaining on the right side of the equal sign (8 gallons
3).
Answer: 24 gallons
Example 2: The price of jellybeans, j, varies directly as the number of pounds, p, that are purchased.
Find the equation that relates the two variables if jellybeans are $1.95 per pound.
(1) y = kx, j = kp
(2) j = 1.95p
Step 1: Remember that the formula for direct variation is: y = kx and substitute the variables from the
question into the appropriate places.
Step 2: Since the jellybeans are always $1.95 per pound, the constant, k, equals 1.95. Substitute 1.95
into the equation for k.
Answer: j = 1.95p
Activities that can help reinforce the concept of direct variation are as follows.
1. Have students solve the equation y = kx for k, and then substitute two sets of (x, y) values into
the equation and compare the values for k. If they are the same, then x and y have a direct variation
relationship.
2. Have the student think of scenarios that show a direct variation relationship. Then, make up
numbers to go with the relationships and have the students practice solving them.
Slope
Slope is the ratio of the vertical difference of two points on a line and the horizontal difference between the
same two points. Slope is also defined as "rise over run" and is found by calculating the difference in the ycoordinates (rise) divided by the difference in the x-coordinates (run). Slope can be found either graphically or
algebraically.
Finding Slope Graphically:
Two points are identified on line z: A (- 3, 8) and B (3, 1).
To find the slope of line z, which passes through points A and B, follow these steps.
Step 1: Start at the left-most point (it is possible to start at either point, but it is important to be
consistent), which is point A (- 3, 8).
Step 2: Count up (positive) or down (negative) until level with the right-most point, point B (3, 1). The
count will be 7 in the downward direction, or - 7. The rise is - 7.
Step 3: Count right (positive) or left (negative) until point B is reached. The count will be 6 to the right,
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or +6. The run is 6.
Step 4: Using the "rise over run" definition of slope, place - 7 on top of the fraction and 6 on the bottom
of the fraction.
Finding Slope Algebraically:
To find the slope of line z algebraically, follow these steps.
Step 1: Use the formula for the slope of a line, where m is the variable that represents slope.
Step 2:
Step 3: Substitute the given coordinate points into the slope formula and simplify the fraction.
It does not matter which method (graphically or algebraically) is used for determining slope.
Regardless, the slope of a line always remains the same.
Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).
Graphically:
Step 1: Start at the left-most point, which is point S (1, 3).
Step 2: Determine the rise (1).
Step 3: Determine the run (1).
Step 4: Using the "rise over run" definition of slope, place 1 on top of the fraction and 1 on the bottom of
the fraction.
Answer: m = 1
Algebraically:
Step 1: Write the formula.
Step 2:
Step 3: Simplify the fraction.
Answer: m = 1
An activity that can reinforce the concept of slope is to have students randomly plot two points on a
coordinate system and then find the slope graphically. They can check their answers by substituting the
two points into the slope formula.
Factoring: Difference of Two Squares
Consider the following equation:
3
4 = 12
The numbers 3 and 4 are said to be factors of the number 12. Factors are terms that can be multiplied
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together to create another term. This concept of factoring is not reserved for numbers, but may be
extended to polynomials as well.
Definitions:
• Factoring is rewriting a mathematical expression as a product of its component factors. One way to
think of factoring is as the opposite or inverse of multiplying.
• Sum means the addition of two terms.
• Difference means the subtraction of two terms.
• Variables are letters that represent unknown numbers.
• Squares are terms whose square roots are whole numbers, zero, or variables whose powers are
divisible by two. For example, 81 is a square because the square root of 81 is 9, and x2 is a square
because the exponent (2) is divisible by two.
• A polynomial is a term or sum/difference of terms. Each term is either a number, a variable, or a
product of a number and one or more variables. Examples: 3x2 + 2x +1; 5x3 ; 4x + 1
• A monomial is a polynomial with one term. Example: 6x2
• A binomial is a polynomial with two terms. Example: 8y4 + 7y
• A trinomial is a polynomial with three terms. Example: 3x2 + 2x +1
The difference of two squares is easily factored using the rule below.
a2 - b2 = (a - b) (a + b)
Example 1: Factor completely.
x2 - 169
Step 1: This problem is a difference of two squares since x2 and 169 are perfect squares. Set up two sets
of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of the
other set.
Step 2: Take the square root of the first term (x2 ) and write x in the first space in each set of parentheses.
Step 3: Take the square root of the second term (169) and write 13 in the second space in each set of
parentheses.
Answer: (x - 13)(x + 13)
NOTE: This format will always work because if (x - 13) and (x + 13) are multiplied together using the
FOIL method, (multiply the First terms, then the Outer terms, then the Inner terms, then the Last terms),
the result is: x2 + 13x - 13x - 169. Combine like terms, or terms that have the same variable(s) with the
same exponent(s), the +13x and -13x terms add to zero. The result is x2 - 169, which is the original
problem.
NOTE: Because of the commutative property of multiplication, the order of the factors can be reversed
and an equivalent expression is made: (x - 13)(x + 13) = (x + 13)(x - 13).
Example 2: Factor completely.
169y2 - 36
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Step 1: This problem is a difference of two squares since 169y2 and 36 are perfect squares. Set up two
sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of
the other set.
Step 2: Take the square root of the first term (169y2 ) and write 13y in the first space in each set of
parentheses.
Step 3: Take the square root of the second term (36) and write 6 in the second space in each set of
parentheses.
Example 3: Factor completely.
2x2 - 50
Step 1: Factor out any common factors. Since 2x2 and 50 can both be evenly divided by 2, perform the
division. NOTE: Don't forget to carry the 2 that was factored out through the entire problem.
Step 2: The remaining problem is a difference of two squares since x2 and 25 are perfect squares. Set up
two sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center
of the other set.
Step 3: Take the square root of the first term (x2 ) and write x in the first space in each set of
parentheses.
Step 4; Take the square root of the second term (25) and write 5 in the second space in each set of
parentheses.
Answer: 2(x - 5)(x + 5)
To help the student further understand this concept, write square numbers on one set of index cards.
Write squared variable terms and/or the product of numbers and variables on another set of index cards.
Have the student draw one card from each set of index cards. Ask the student to factor the differences
between each pair of cards. He or she will have to find the square roots of the terms on the cards to
correctly complete the task.
Radicals: Addition/Subtraction - A
A radical sign looks like a check mark with a line across the top. The radical sign is used to
communicate square roots. The square root of a number is the number that when squared, or multiplied
by itself, is equal to the radicand. The radicand is the number under the radical sign. For example, the
square root of 64 is equal to 8, because 8 squared (8
8) is 64, and 64 is the radicand.
Students must be able to combine (add and subtract) radicals. In order for radicals to be combined, the
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radicands must be identical. The square root of 3 and the square root of 7 cannot be added together in
their present form because their radicands (in this case, 3 and 7) are not identical. When combining
radicals, the radicands themselves do not get added or subtracted.
Example 1: Add and simplify.
Step 1: Check to see if the radicands are identical. In this case, both radicands are 5, so the radicals can
be added.
Step 2: Add the coefficients (numbers in front of the radicals). The answer is the sum of the coefficients
times the radical.
Example 2: Subtract and simplify.
Step 1: Check to see if the radicands are identical. In this case, both radicands are 3, so the radicals can
be subtracted.
Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical.
There is a silent 1 in front of the radical in the final answer.
Example 3: Simplify.
Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 7, so those two
radicals can be subtracted.
Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical,
plus the remaining radical.
Example 4: Simplify.
Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 3, so these two
radicals can be added. Add the coefficients and then multiply the sum by the radical.
Step 2: Rewrite radical 16 in simplest form. Since 16 = 4
4, radical 16 can be simplified to 4.
Step 3: The final answer is the answer to step 1 minus the answer to step 2.
An activity that can help reinforce the concept of adding and subtracting radicals is to name radicals
with the names of animals. For example, name the square root of 6 "elephant," and the square root of 5
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"lion." Then say, "If you have 5 elephants and you add 10 more elephants and 2 lions, how many
elephants and lions do you have altogether?" Answer: 15 elephants and 2 lions, or, 15 times the square
root of 6 plus 2 times the square root of 5. This activity keeps students from trying to combine radicands
in the answer.
Inequalities - C
An inequality is a number sentence that uses "is greater than," "is greater than or equal to," "is less than," "is
less than or equal to," or "is not equal to" symbols. For example, 6n > 4 is a number sentence with an inequality
symbol, and is read "six times n is greater than four." Inequalities can be identified in real world situations by
expressions such as "is less than," "is more than," "at least," and "at most." Sentences can be translated to
number sentences using the following symbols.
An inequality can be solved for a variable (a letter that represents a number) in the same way that an
equation is solved. An example of solving an inequality follows.
Example 1: Solve the inequality for x.
2x + 7 > 11
Step 1: Subtract 7 from each side of the inequality symbol to isolate the variable term, 2x, and simplify.
Step 2: Divide both sides of the inequality by 2 to completely isolate x.
Answer: x > 2
A compound inequality has more than one condition and can be identified in a number sentence by two
inequality symbols, or in a real world application by the words "and" or "or." For example, 3 < x < 9 is
a compound inequality. To read a compound inequality, start in the center and read left: x > 3, then go
back to the center and read to the right: x < 9. The compound inequality is a combination of x > 3 AND
x < 9. Solving a compound inequality is very similar to solving a single inequality.
Example 2: Solve the compound inequality for x.
4 < 3x + 1 < 7
Step 1: Subtract 1 from the left, center, and right side of the number sentence to isolate the variable term,
3x.
Step 2: Divide the left, center, and right side of the number sentence by 3 to isolate x.
Answer: 1 < x < 2. This answer can be interpreted to mean that x is greater than 1 and x is less than 2.
Although the above examples do not include negative numbers, students must remember to switch the
inequality sign when multiplying or dividing by a negative number.
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Once the student is familiar with solving compound inequalities, he or she should be ready to solve them
in real world situations.
Example 3: The telephone company charges $21.95 per month for basic service plus $0.17 per local
call. What is the maximum number of calls that can be made in a month if a family can spend at least
$27.50 and at most $32.75 per month for telephone service? Round your answer to the nearest whole
number of calls that satisfies the inequality.
Step 1: Translate the information in the problem to a compound inequality. Let c represent the number
of calls. The family gets charged $21.95 + $0.17c each month (the service charge plus local calls).
Since the charges are at least $27.50 and at most $32.75, place the minimum amount of $27.50 to the
left of the expression that represents the charges and the maximum amount of $32.75 to the right of the
expression. Refer to the chart above to see which inequality symbol to use. Since the question uses the
words at least and at most, the
needs to be used. Place the
on either side of the charges
expression.
Step 2: Isolate the variable term $0.17c by subtracting $21.95 from the left, center, and right of the
compound inequality. Simplify.
Step 3: Divide each term of the new compound inequality by $0.17 to isolate c.
Step 4: Simplify the inequality.
Step 5: Since the maximum number of calls must be less than or equal to 63.53, the whole number that
will satisfy the inequality is 63. Rounding up to 64 would no longer satisfy the inequality because 64 is
greater than 63.53.
Answer: 63 calls
Note: If the question had asked what the minimum number of calls would be, the answer would be 33,
since the minimum number of calls has to be greater than or equal to 32.65.
An activity that can help reinforce the concept of inequalities is to ask the student to make a budget
using the amount of money he or she earns each month (approximate or make up an amount if
necessary). Then, make up scenarios using the budget. For example:
It costs $120 per month to insure a car, plus $2.05 per gallon of gasoline used. If you have budgeted at
least $150.00 and at most $200.00 for car expenses this month, what is the maximum number of gallons
of gas you can buy?
Function Rules
A function rule is an equation relating two variables. Variables are letters that represent unknown
numbers, such as x and y.
In a function, the value of the output variable, y, depends on the value of the input variable, x. The
function rule takes a given value of x, and performs an operation (addition, subtraction, multiplication,
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or division) on it to make it equal to y.
[i]
[i]
A table of related values for input-output variables that can be generated from a function rule is called a
function table. It is important for students to understand what a function table represents. To create the
function table below, x-values were chosen and placed in the table. Next, the x-values were substituted
into the function rule and a y-value was generated for each x-value. The final function table appears on
the right without the function rule.
It is possible to determine the function rule from a table of values for related input-output variables. The
first two examples deal with simple function rules that use addition and subtraction only.
Example 1: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: Determine what needs to be added to or subtracted from - 3 to make it equal to 3. The solution is,
"add 6," because - 3 + 6 = 3.
Step 2: Add 6 to each of the remaining x-values, and verify that they match the values listed for y.
Step 3: Write the function rule: y = x + 6.
Example 2: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: Determine what needs to be added to or subtracted from 1 to make it equal to - 4. The solution is,
"subtract 5," because 1 - 5 = - 4.
Step 2: Subtract 5 from each of the remaining x-values, and verify that they match the values listed for y.
Step 3: Write the function rule: y = x - 5.
Function rules can also be of the form y = ax + b, or y = ax. Rules of this form require the student to use
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multiplication, as well as addition and subtraction.
Example 3: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: First, the student should test the y values to see if they are multiples of the x values. If the
function is of the form y = ax, then the student should be able to multiply all of the values in the x
column by the same number, and arrive at the values in the y column. Multiplying - 4 by 3 yields - 12.
Step 2: Multiply the remaining x values by 3, and verify that they match the values listed for y.
Step 3: Write the function rule: y = 3x.
Example 4: What function rule relates the values of the input variable, x, to the values of the output
variable, y, in the table below?
Step 1: First, the student should test the y values to see if they are multiples of the x values. If the
function is of the form y = ax, then the student should be able to multiply all of the values in the x
column by the same number, and arrive at the values in the y column. If the student is unable to
determine a single factor that yields the output values, he or she should continue to step 2. Since there
isn't a factor that yields - 19 when multiplied by - 4, the student should continue to step 2.
Step 2: If there is an input value equal to 0, the student should always begin with this value. This line in
the table will help the student to determine the value that is being added or subtracted since any factor
multiplied by 0 equals zero. In this case, 3 is being subtracted from x because 0 - 3 = - 3. Therefore, the
function must be of the form y = ax - 3.
Step 3: The student must now determine what value of a satisfies the function. Determining this value is
made easier by performing the inverse of the operation found in step 2 on a chosen y value. Adding 3 to
-19 yields - 16. Since - 4, (the corresponding x value) must be multiply by 4 in order to yield -16, the
factor 4 is the correct value for a.
Step 4: Multiply each input value by 4 and subtract 3 in order to verify that they match the values listed
for y.
Step 5: Write the function rule: y = 4x - 3.
Exponential Functions
An exponential function is an equation that has a variable in the exponent. The functions below are all
exponential functions, where f(x) represents the letter y and is dependent on the value of x.
When x-values are substituted into the equation and the right side of the equal sign is simplified, a list of
f(x) (or y) values is obtained. This process gives you a set of (x, y) coordinate pairs that can be plotted
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(or graphed) on a coordinate plane.
Step 1: Make a table of values with the following column headings.
Step 2: Choose values to enter into the x column (remember to
choose the smallest possible values to keep computations as simple as possible). It is generally
necessary for the values to contain both negative and positive numbers and the number zero. As few as
three values could be used, but this example will use these five values: -2, -1, 0, 1, 2. Enter the values
into the table.
Step 3: Substitute each x-value into the equation in the center column for x and simplify. This will
determine the y-values for the coordinate points.
Step 4: Write the (x, y) pairs generated in Step 3 into the (x, y)
column, using the values from the x column and the y-values from the center column.
Step 5: Plot the points on a coordinate plane to graph the
exponential function.
Step 6: Compare this graph to the solution choices and choose the correct
answer, which is choice D.
Answer: D.
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Step 1: Make a table of values and choose x-values to substitute into the equation.
Step 2: Substitute the values in the x column into the equation in
the center column to determine the corresponding y-values. Write the coordinate points in the (x, y)
column.
Step 3: Plot the points on a coordinate plane to graph the
exponential function.
Step 4: Compare this graph to the curves in the question and choose the
correct answer.
Answer: W
Comparing Graphs of Exponential Functions:
The standard form for an exponential function is shown below.
For purposes of comparing graphs of exponential functions, two more
variables need to be added, such that the standard form becomes the form shown below.
The value of c determines whether the graph shifts upward or downward
and the value of d determines whether the graph shifts right or left. See the table below.
Example 3:
Step 1:
Step 2: Look at the table above to determine that the graph of the exponential
function should move upward 3 units and to the right 4 units.
Step 3: Compare the actual graphs below to see if these predictions were right.
An activity that will help reinforce the skill of graphing exponential functions is to explore exponential
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functions with the student using a graphing calculator. Make up a list of exponential functions like the
one below.
Follow the steps to graph each function using a graphing calculator. Notice the way changing a sign or
adding a term alters the graph.
Step 1: Follow the instructions in your graphing calculator manual to set the viewing window to the
following values: x min = -10, x max = 10, xscl = 1; y min = -10, y max = 10, y scl = 1. Exit the
window.
Step 2: Press the y = key to open the equation editor. Delete any equations that are listed. Enter the
following keystrokes for the first exponential equation on the list: 5^x. Press Enter.
Step 3: Press the Graph key to display the graph of the function.
**NOTE: This activity could be completed using graph paper and tables of values to graph the
exponential functions.
Exponential Notation - F
In the expression 32 , the number 2 is called an exponent and the number 3 is called a base. The exponent
determines the number of times the base is multiplied by itself. For example: 32 = (3)(3) = 9.
Rational exponents are exponents that are in fraction form. The following are examples of expressions
with rational exponents.
Expressions with rational exponents can be written in radical form. The numerator is the number of
times the base, which is placed under the radical sign, is multiplied by itself. The denominator is the root
of the radical expression. A root is the inverse of an exponent. The examples below show how
expressions with rational exponents are written in radical form.
A root of 2 is called a square root and is the inverse of the exponent 2. A root of 3 is called a cube root
and is the inverse of the exponent 3. The following examples show how to simplify expressions with
rational exponents in two ways.
Simplifying by converting to radical form:
Example 1: Simplify.
Step 1: Rewrite the rational expression in radical form.
Step 2: Simplify: 8 to the first power equals 8.
Step 3: Factor 8 into (2)(2)(2).
Step 4: Since there are three factors of 2 under the cube root, the radical expression will simplify to 2.
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Answer: 2
Simplifying using exponent laws:
Step 1: Rewrite the expression and factor 8 into (2)(2)(2).
Step 2: Write (2)(2)(2) in exponential form.
Step 3: Multiply 3 and 1/3 (remember that when an exponent is taken to another power, the two powers
are multiplied) to get 1. Simplify 2 to the first power to 2.
Answer: 2
Example 2: Simplify.
Step 1: Rewrite the expression as a radical expression.
Step 2: Simplify 322 : 32
32 = 1,024.
Step 3: Factor 1,024 into (4)(4)(4)(4)(4). Always try to factor the number under the radical into the
same number of factors as the root.
Step 4: Since this is a 5th root and there are five 4s under the radical, the radical expression will simplify
to 4.
Answer: 4
Activities that can help reinforce the concept of rational exponents are as follows.
1. Take 30 index cards, and make three piles of ten cards each. Write expressions with rational
exponents on the first set of ten cards. On the second set, write the radical form for each expression you
wrote on the first ten cards. On the third set, write the simplified answer to the ten expressions. Shuffle
all of the cards and lay them face down on a table. Each player takes turns flipping three cards. If all
three cards go together, the player has made a match and gets to try again. If not, it is the next player's
turn.
2. Name expressions with rational roots, and ask the student to factor the base into the same number of
identical factors as the root. For example, if the root is 1/6, the student should factor the base into 6
identical factors, so that when the factors are rewritten in exponential form, and the exponent is
multiplied by the root, the product is 1.
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Polynomials: Subtraction
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial.
The sum or difference of two or more monomials is called a polynomial.
Here is an example of a polynomial:
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
To subtract polynomials, first write the polynomials as one long polynomial. Then distribute the
subtraction sign through the second polynomial. Finally, combine like terms. Practice by subtracting the
following polynomials.
Example 1:
Step 1: Set up the two polynomials as one long polynomial. Since the problem is to subtract one
polynomial from another, the second polynomial in the problem must be written first.
Step 2: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 3: Rewrite the polynomial after changing the signs in the second polynomial.
Step 4: Combine like terms.
Answer: 5p + 9
Example 2: Subtract four times a number decreased by ten from eight times the same number less six.
Step 1: "Four times a number decreased by ten" can be written (4x - 10).
Step 2: "Eight times the same number less six" can be written (8x - 6).
Step 3: Now the problem reads: Subtract (4x - 10) from (8x - 6).
Step 4: Set up the polynomials as one long polynomial.
Step 5: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 6: Rewrite the entire polynomial after changing the signs in the second polynomial.
Step 7: Combine like terms.
(4x - 10) subtracted from (8x - 6) equals 4x + 4.
Answer: 4x + 4
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Example 3: Find area of the shaded region.
Step 1: Determine the area of the large rectangle by multiplying the length (2x + 7) by the width (3x).
This involves multiplying each term in (2x + 7) by 3x.
Step 2: Determine the area of the small rectangle by multiplying the length (2x - 2) by the width (x).
This involves multiplying each term in (2x - 2) by x.
Step 3: Now subtract the area of the small rectangle from the area of the large rectangle. Remember to
put the second polynomial in parentheses since this is subtraction.
Step 4: Distribute the subtraction sign through the second polynomial. This involves changing the sign
of each term in the second polynomial.
Step 5: Combine like terms.
Answer:
Equations: Order of Operations
An equation is a statement in which two numbers or two expressions are set equal to each other. For example,
5 + 3 = 8 and 16 = 3c + 4 are equations.
When solving equations, find the value of the variable by getting the variable alone on one side of the
equal sign. To do this, undo any operations on the variable by using the inverse operation. Any operation
done on one side of the equal sign must be done on the other side of the equal sign in order to keep the
statement true.
If a number has been added to the variable, subtract the number from both sides of the equation.
m + 3 = 5
- 3 - 3
m = 2
If a number has been subtracted from the variable, add the number to both sides of the equation.
b - 7 = 9
+ 7 + 7
b = 16
If a variable has been multiplied by a nonzero number, divide both sides by the number.
If a variable has been divided by a number, multiply both sides by the number.
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When solving 2-step equations, we must first undo the addition or subtraction using the inverse
operation, then undo the multiplication or division:
Example 1: Solve the equation for t.
5(t - 4) = t + 12
Step 1: Multiply 5 times the terms inside the parenthesis.
Step 2: Add 20 to both sides of the equation
Step 3: Subtract t from both sides of the equation
Step 4: Divide both sides of the equation by 4.
Answer: t = 8
Example 2: Evaluate the expression for c = 3:
2(c + 4) + 2(15)
(1) 2(3 + 4) + 2(15)
(2) 2(7) + 2(15)
(3) 14 + 30
(4) 44
Step 1: Substitute 3 in place of 'c' in the expression.
Step 2: Add the numbers in parentheses.
Step 3: Rewrite the equation after performing all multiplications in order from left to right.
Step 4: Add 14 and 30 to get 44.
Answer: 44
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Functions: Linear
This skill requires students to solve real world problems involving two linear functions. A linear function is a
function whose graph forms a non-vertical straight line.
Story problems are often very difficult for students to master. It may be beneficial first to confirm that
the student is comfortable with solving two linear equations outside of a real world context. To do this,
students must review how to solve linear equations. In some variable equations, students are given the
value of one variable. Letters expressing unknown values in equations are called variables. To find the
value of the second variable, substitution must be used.
If y = 4x, and x = 9, substitute the given value of x in the equation, y = 4x. The result is y = 4(9).
Calculate the right side of the equation to get y = 36. Now the values of both variables are known: x = 9
and y = 36.
To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each
side of the equation by 4. The result is x = 9, so the values for x and y are both correct.
Here is another situation where the value of a variable is substituted into an equation to solve for x.
Example 1:
Solve for x.
x = 3y + 9
y=-2
Solution: Substitute - 2 in place of y in the first equation. Then, solve for x.
x = 3(- 2) + 9
x=-6+9
x=3
Answer: x = 3
This concept can be used to solve two equations that include the same two variables.
Example 2:
Solve for x and y in the following equations.
y = 2x - 3
3x = y + 10
Solution:
If y = 2x - 3 and 3x = y + 10, there is not a number to substitute in for either of the variables. So, an
expression must be substituted for a variable. In this case, the y variable is already isolated. It is equal
to 2x - 3. This can be substituted in for y in the second equation.
(1) 3x = y + 10
3x = (2x - 3) + 10
(2) 3x = 2x + (- 3 + 10)
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3x = 2x + 7
(3) 3x = 2x + 7
-2x -2x
x=7
(4) y = 2x - 3
y = 2(7) - 3
(5) y = 14 - 3
y = 11
Step 1: Substitute 2x - 3 for y in the second equation.
Step 2: Simplify the equation by removing the parentheses and combining the like terms - 3 and 10.
Step 3: Isolate the variable by subtracting 2x from each side and combining like terms.
Step 4: Since there is now a value for one of the variables, it can be substituted back into either of the
original equations. In this example, the first equation was used.
Step 5: Calculate the right side of the equation to get y = 11.
Answer: x = 7 and y = 11
Several questions from this skill relate to measurement concepts. The formula for perimeter is
occasionally needed to find solutions to story problems. Perimeter is the measurement around a figure.
Example 3:
Find the perimeter for the figure below.
Solution:To calculate the perimeter of a figure, add the lengths of all the sides of the figure. For
example, this figure has four sides measuring 3 inches, 7 inches, 3 inches, and 7 inches.
P = 7 + 3 + 7 + 3 = 20
Answer: The perimeter of this figure is 20 inches.
Story problems that involve perimeter and two linear equations can be solved by writing two equations,
then using the substitution method that was reviewed earlier.
Example 4:
Emilio built a square sandbox for his dog with a perimeter of 15 m. Since his dog has grown, he wants
to rebuild the sandbox so that each side is 2 m longer than the original sandbox. Before he buys the
materials to build the new sandbox, he needs to know the perimeter. What is the perimeter of the new
sandbox?
Solution:
To solve this problem it is probably best to start with drawing a picture. Because we do not know the
length of the sides of the old enclosure, they can be assigned a variable, x. The new enclosure must have
sides that are 2 m longer than the original, so the sides can be written as x + 2.
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(1) For the old enclosure, x + x + x + x = 15, or 4x = 15
For the new enclosure, (x + 2) + (x + 2) + (x + 2) + (x + 2) = y, or 4(x + 2) = y.
(2) 4x = 15, so x = 3.75
(3) 4(3.75 + 2) = y
(4) y = 23
Step 1: An equation can be written for the perimeter of the new and old enclosures based on the
information in the problem. For the new enclosure, we do not know the perimeter so we give it a
different variable, y.
Step 2: The x must be isolated in the first equation so that it can be substituted into the second.
Step 3: This value can be placed in the second equation for x.
Step 4: Calculate.
The perimeter of the new enclosure will be 23 m.
Answer: 23 m
Another common application of finding the solution to two equations in solving real world problems
involves money.
Example 5: Ariel wants to go snowboarding this weekend. Ice Mountain charges $24 for a lift ticket
and $8 per hour of snowboarding. Flurry Ridge charges $29 for a lift ticket and $6 per hour of
snowboarding. Ariel must get a lift ticket to snowboard and she wants to snowboard for 5 hours. Which
ski area would be cheaper for Ariel's snowboarding trip?
(1) For Ice Mountain, the cost = 24 + 8h.
For Flurry Ridge, the cost = 29 + 6h.
(2) For Ice Mountain, 24 + 8(5) = 64
For Flurry Ridge, 29 + 6(5) = 59.
Step 1: Set up two equations, one for each snowboarding location. We can use the variable h for the
number of hours.
Step 2: Since Ariel wants to snowboard for 5 hours, we can substitute 5 in for h in each of the equations
and solve.
Answer: Ice Mountain will cost her $64 and Flurry Ridge costs her only $59, so Flurry Ridge is
cheaper.
To help students understand the application of linear equations in solving real world problems, students
can find the costs per unit of time or distance of different services with the help of a telephone book, the
newspaper, or the internet. If specific cost information is not listed, call the company for information.
For example, a car rental company may charge a base fee of $35 per day and $0.59 per mile or a
catering company may charge a rental fee of $100 for materials and then $60 per hour for waitstaff
services. Then, have students calculate which would be the cheapest and which would be the most
expensive for a specified distance a rental car may drive or amount of time a catered party may last.
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Quadratic Equations: Real World Problems
A quadratic equation is a second-order polynomial equation in a single variable. Second-order means that one
of the terms contains the variable to the power of two and no other exponents are greater than two. Students
will apply quadratic equations to evaluate real world problems in this question set.
Real world problems are often difficult for students to master. It may be beneficial first to confirm that
the student is comfortable with quadratic equations outside of a real world context. To do this, students
must first review the basic principles of quadratic equations, including how to distribute terms in order
to obtain the equation in the proper form, how to factor, and eventually, how to solve for the variable.
Review of Definitions:
• A variable is an unknown number in an equation and is usually represented by a letter such as x.
• A coefficient is a number representing a fixed, or constant, value that is multiplied with a variable.
• A polynomial is a term or sum of terms. Each term is either a number or the product of a number and
one or more variables. For example, 2x2 + 3 is a polynomial with two terms, 2x2 and 3.
• A binomial is a polynomial with two terms, and a trinomial is a polynomial with three terms.
• Factoring is the breaking up of quantities into products of their component factors. One way to think
of factoring is as the opposite, or inverse, of multiplying.
Quadratic functions are written in the form:
where f(x) is read "f of x." The letters a, b, and c are constants and a
0.
When solving quadratic equations, replace the y or f(x) with zero. It is important to remember that all
quadratic equations have two answers. However, the answers could be imaginary (contain a square root
of a negative number), positive, or negative. When dealing with real world problems, the positive
answer is typically the only one that would make sense in the context and would therefore be the only
correct answer.
Below are a few examples of quadratic equations:
Sometimes, quadratic equations come in a form where they have already been factored. Some examples
of these equations are shown below.
(x + 5)(x - 1) = 7
or
2x(x + 6) = 54
In order to find the solution to quadratic equations, they have to be set equal to zero. To get the
equations in this form, multiply out the terms to eliminate the parentheses and subtract appropriately
until the polynomial is equal to zero. Once an equation is written in the proper form, the next step
towards solving for the variable is factoring the equation.
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Example 1:
Factor completely and solve for y.
y2 + 4y - 2 = 3
Step 1: Subtract 3 from both sides of the equation so that the equation is equal to zero.
Step 2: Determine the factors. Since the coefficient of the y2 term is equal to 1, focus on the last term,
in this case, - 5. If factors of - 5 can be found that add up to the coefficient of the "y" term (in this
example, 4), the trinomial can be factored. The factors of - 5 are 5 and - 1 or - 5 and 1. When we add
the first pair of numbers, 5 and - 1, we get 4, the number we are seeking.
Step 3: Check your answer. To check the result, multiply the binomial factors together. Binomials are
multiplied using the FOIL method. FOIL is the abbreviation for First, Outside, Inside, and Last.
• Multiply the First terms of each binomial (y
• Multiply the Outside terms of each binomial (y
• Multiply the Inside terms of each binomial (5
y = y2 )
-1 = -1y)
y = 5y)
• Multiply the Last terms of each binomial (5
- 1 = - 5)
• Combine like terms to simplify the polynomial.
Step 4: Set each factor equal to zero. When we multiply any value by zero the result is zero. Therefore,
if either factor is equal to zero, the entire expression is equal to zero.
Step 5: Solve each of these equations by isolating the variables.
Answer: y = - 5 and y = 1
A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type
are easily factored once the pattern is remembered.
Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6,
respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6).
Notice that if these terms are multiplied together using the appropriate FOIL method, the original
polynomial is formed. Polynomials that are in the "difference of squares" form, that is, a2 - b2 , can
always be factored as the sum of the square roots, (a + b), multiplied by the difference of the square
roots, (a - b).
Using Quadratic Equations in Real World Problems:
When solving real world problems, clues from the text of the problem must be used to set up the
equations. Quadratic equations are often used to find solutions to story problems. The story problems
in this set also occasionally require the use of formulas for the area of a rectangle, the area of a triangle,
Page 22
or the Pythagorean Theorem.
Area is the measure of the interior of a two-dimensional region and is calculated in square units. The
formulas used to find the area of a rectangle and a triangle are shown below.
Area of a rectangle = length x width, or A = lw. A square is a rectangle with four equal sides, so the
formula for the area of a square is often shown as Area of a square = side2 , or A = s2 because the length
is equal to the width for all squares.
Area of a triangle =1/2 x base x height, or A = (1/2)bh. The height of a triangle is the length of the
line segment that extends from the vertex opposite the base and is perpendicular to the base, as shown
below.
The Pythagorean theorem is used to find the lengths of the sides and hypotenuse of a right triangle. The
Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of
the squares of the legs of that triangle. A right triangle is a triangle containing a 90º angle, the
hypotenuse is the side opposite the right angle (longest side of the triangle), and the legs are the two
other sides.
Using the diagram above, it is important to remember that a and b always denote legs of a right triangle
and c always denotes the hypotenuse of a right triangle.
Sometimes, the length or width are unknown in an area problem or the leg is unknown in a right triangle
problem. Examples of these types of problems are provided below.
Example 2:
Colleen is swimming in the ocean. She swam 4 miles south and then turned and swam 3 miles west. If
she turns and swims straight back to the point where she started swimming, how many more miles does
she need to swim?
Step 1: Determine the values of a, b, and c. In this case, a and b are known and c is unknown.
Step 2: Substitute the values of a and b into the Pythagorean theorem.
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Step 3: Square 3 (3
3 = 9) and square 4 (4
4 = 16).
Step 4: Add 9 and 16 to get 25.
Step 5: Take the square root of each side of the equation. The square root of 25 is 5 and the square root
of c2 is c.
Answer: She will swim 5 miles.
Example 3:
Find the height of a triangular sail that has an area of 60 m2 and a base of 8 m.
Step 1:
Step 2:
Step 3:
Step 4:
Begin with the formula for the area of a triangle.
Substitute in 60 for the area and 8 for the base.
Simplify the equation by multiplying 1/2 and 8 to get 4.
Isolate the height by dividing each side of the equation by 4.
Answer: The height of the sail is 15 m.
When working with story problems, use a variable for the unknown value and clues within the wording
to write the equations that will need to be solved. For example, if a problem reads "the length is 4 m
longer than the width," then the length can be represented by w + 4. After an entire equation is formed,
it can then be solved.
Example 4:
The width of a rectangular cake pan is 3 inches shorter than the length of the pan. The area of the cake
pan is 108 in.2 What is the length of the cake pan?
Solution:
(1)
(2) x(x - 3) = 108
(3) (x)(x) - (x)(3) = 108
x2 - 3x = 108
(4) x2 - 3x - 108 = 0
(5) (x - 12)(x + 9) = 0
(6) x - 12 = 0 and x + 9 = 0
(7) x = 12 and x = - 9
Step 1: Drawing a picture of the cake pan with the information in the problem may help in solving the
problem. The length is the unknown value, so it can be represented by the variable x. The width is 3
inches shorter than the length, so it can be written as x - 3.
Step 2: Since this problem deals with the area, length, and width of a rectangle, we must use the
formula for the area of a rectangle, Area = length x width, and substitute in x for the length, x - 3 for the
width, and 108 in.2 for the area.
Step 3: Distribute the x and simplify.
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Step 4:
Step 5:
Step 6:
Step 7:
Subtract 108 from both sides to put it in the proper form.
Factor the equation.
Set each polynomial equal to zero.
Isolate the variable in each problem.
Answer: The answer x = - 9 does not make sense because the length of a cake pan cannot be a negative
value. That answer is discarded and the correct answer must be x = 12. The length of the cake pan is 12
in.
To help students understand the application of quadratic equations in solving real world problems, help
the student find the missing dimensions of various objects around the house when given the area and
information about one of the dimensions. For example: Measure a picture frame, door, etc., calculate
the area of the object, and create a relationship between the dimensions (the length of the door is twice
the width). Then have the student find the missing dimension when given the area and the relationship
between the dimensions. An activity to see how the Pythagorean theorem works is to lean a board or
ladder against a wall to create a right triangle. Measure two dimensions of the triangle and use the
Pythagorean theorem to determine the length of the third side of the triangle.
Functions/Relations - A
A relation is a set of ordered pairs that represent a relationship between the elements of the two sets. A function
is a special type of relation, where each element of the first set (x-values) corresponds to an unique element of
the second set (y-values). The first set of numbers is commonly known as the input and the second set as the
output. The input, or x-values, are entered into the equation. Once evaluated, the result is the output, or yvalues. In other words, in order for a relation to be a function, for each x-value there can be no more than one
value of y. Some examples of relations are given below, with input values in A mapped to output values in B.
Relations 1 and 2 are functions, while relation 3 is not a function. The input value - 1 in relation 3 is
matched to more than one output value (3 and 5), so the relation is not a function.
Example 1:
Which of the following relations is not a function?
Solution:
If there is a value of x resulting in more than one value of y, the relation is not a function. This only
occurs in the third set of numbers with (9, - 1) and (9, 4). Therefore, set C is not a function.
Answer: Set C is not a function.
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Example 2:
Which of the following points, if removed from the set, would make the set a function?
Solution:
The ordered pairs (- 4, 5) and (- 4, 4) have the same x values but different y values. Therefore, if either
point is removed from the set, the remaining ordered pairs will represent a function.
Answer: Remove either (- 4, 5) or (- 4, 4).
Equations With Two Variables
This study guide will focus on solving equations that contain two variables.
Remember:
Variables are letters or symbols that represent numbers that are unknown.
Expressions are variables or combinations of variables, numbers, and symbols that represent a
mathematical relationship. Expressions do not have equal signs, but can be evaluated or simplified.
Example: y - 9
Equations are expressions that contain equal signs. They can be solved, but not evaluated.
Example: 5 + x = 13
Like terms are those terms that have the same variable(s) in common, or no variable. Like terms can be
combined.
Examples: 2 and 3 are like terms, 2x and 3x are like terms, but 2 and 3x are not like terms.
Solving Equations
To solve an equation, it is necessary to "undo" what was done to the variable in question. Another way
to think about this is to do the order of operations in reverse.
P = Parenthesis, E = Exponents, M = Multiplication, D = Division, A = Addition, S = Subtraction.
**In the normal direction, first evaluate multiplication and division from left to right (whichever comes
first), then evaluate addition and subtraction from left to right (whichever comes first).
**To "undo" what was done, reverse the direction.
•Addition "undoes" subtraction.
•Subtraction "undoes" addition.
•Multiplication "undoes" division.
•Division "undoes" multiplication.
Addition/Subtraction Properties - Adding or subtracting the same real number to each side of an
equation will result in an equivalent equation.
Example 1: Solve 16 = 5 + x - 8 for x.
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Step 1: Begin to isolate the variable, x, by adding 8 to both sides of the equation. Remember to only add
and subtract like terms, so the 8 must be added to the - 8 and the 16.
Step 2: Subtract 5 from both sides of the equation to completely isolate the variable, x.
Answer: x = 19
Multiplication/Division Properties - Multiplying or dividing each side of an equation by the same
(nonzero) number will result in an equivalent equation.
Example 2: Solve the following equation for t.
Step 1: Multiply both sides of the equation by 17 to eliminate the fraction.
Step 2: Divide both sides of the equation by 4 to isolate the variable, t.
Answer: t =136.
Solving Equations With Two Variables
In order to solve equations that contain two variables, the student will need to solve for one variable in
terms of the other. This means that the answer may not be entirely numeric.
The same properties as described above should be used when solving for a given variable in a twovariable equation. Students should treat the second variable as if it were a number.
Example 3: Solve 30w - 9v = 6 for w.
Step 1: Begin to isolate w by adding 9v to both sides of the equation. Remember to only combine like
terms.
Step 2: Completely isolate the variable, w, by dividing by 30 on both sides of the equation.
Step 3: Simplify the solution. Since 6, 9, and 30 can all be divided by 3, divide the numerator and
denominator of the fraction by 3.
Example 4: Solve the following equation for r.
Step 1: Begin to isolate r by adding 7 to both sides of the equation. Remember to only combine like
terms. 10s + 7 cannot be combined because they are not like terms.
Step 2: Completely isolate r by multiplying both sides of the equation by 7.
Answer: r = 7(10s +7)
Example 5: Solve the following equation for a.
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5(a - b) = 14b
Step 1: Questions of this type involve using the distributive property. The distributive property states
that for all numbers a, b, and c, a(b + c) = ab + ac. Therefore, begin by multiplying 5 by a and by b.
Step 2: Next, begin to isolate a by adding 5b to both sides of the equation. 14b and 5b are considered
like terms so they can be added together to get 19b.
Step 3: Completely isolate the variable, a, by dividing both sides of the equation by 5.
Example 6: Solve the following equation for x.
8xy - 9x = 20
Step 1: Questions of this type involve using the distributive property to write x as a factor. Begin by
factoring out the x as a common factor.
Step 2: Isolate the variable, x, by dividing both sides of the equation by (8y - 9).
Ratio/Proportion - C
A ratio is a comparison of two numbers expressed as a quotient. Ratios can be written in three ways: a fraction
(3/5), a ratio (3:5), or a phrase (3 to 5). Like fractions, ratios refer to a specific comparison. The ratios 3/5, 3:5,
and 3 to 5 (as in "the ratio of cellos to violins was 3 to 5") all express the same ratio or comparison. A
proportion reflects the equivalency of two ratios. The ratio 3/5 expresses the same proportion as the ratio 15/25.
To understand how ratios operate, students need to understand equivalent fractions. Fractions represent
portions or parts. For every fraction, there is a corresponding portion. The fraction 1/2 communicates
one portion out of two, but this specific portion can also be communicated by the fractions 2/4, 3/6,
8/16, 10/20, etc. All of these fractions are equal to 1/2 because the relationship between the numerator
and denominator in 1/2 is the same relationship between the numerators and denominators in 2/4, 3/6,
8/16, and 10/20. Ratios and proportions operate in a similar manner. The ratio 2:5 communicates a
specific portion. The ratio 4:10 communicates the same portion because 4:10 reduces to 2:5.
Example 1: Sandra has 15 lollipops and 25 jellybeans. What is the ratio of lollipops to jellybeans?
Answer: There are 15 lollipops to 25 jellybeans, so the ratio of lollipops to jellybeans is 15:25.
Proportions occur when two ratios are equal. In a proportion the cross products of the terms are equal.
Example 2: Is the following proportion True or False?
1/3 = 3/9
The cross products are both equal to 9, so the proportion is TRUE. If the cross products were not equal,
the proportion would be FALSE.
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Sometimes you must find the value of a variable in a proportion. To solve the proportion, you must find
the value of the variable that makes both ratios equal.
Example 3: What is the value of a?
9/12 = a/48
Step 1: Find the cross products. Multiply 48 by 9 and 12 by 'a'.
Step 2: 48 x 9 = 432 and 12 x a = 12a. Rewrite the equation with the new products.
Step 3: Divide each side of the equation by 12 to isolate the variable 'a'.
Step 4: Divide 432 by 12 to get a = 36.
Answer: a = 36
Example 4: Apples are 13 for $3.35. What will 32 apples cost? (Round answer to the nearest
hundredth).
Step 1: Write the appropriate proportion. Let b represent the cost of 32 apples.
Step 2: Write the cross products. Multiply b by 13 and multiply 3.35 by 32.
Step 3: Rewrite the equation with the new values.
Step 4: Divide each side of the equation by 13 to isolate the b.
Step 5: 107.20 ÷ 13 = 8.246153846 We only need to write 3 decimal places because we are going to
round to the hundredths place (the second decimal place). The three dots above represent that the
decimal number continues on past that point.
Step 6: Round 8.246 to the hundredth place. The 6 tells us to round the 4 up to a 5. 8.246 approximately
equals 8.25.
Answer: $8.25
Take a look at the solution. Does it make sense that 32 apples would cost $8.25? We know that 32
apples would cost more than twice as much as 13 apples (13 x 2 = 26), but less that three times as much
(13 x 3 = 39). The solution is more than twice as much as $3.35 (2 x $3.35 = $6.70), but less than three
times as much (3 x $3.35 = 10.05), so our answer makes sense.
Example 5: Rachel scored 25% of the team points. If the team scored 40 points, how many did Rachel
score?
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Step 1: Write a proportion to represent the problem.
Step 2: Cross multiply. This involves multiplying the bottom of one ratio by the top of the other and vice
versa. Perform the multiplication.
Step 3: Divide both sides of the equation by 100.
Answer: 10 points
Function/Pattern - C
Students must identify patterns in numbers (such as missing elements) and shapes, and then draw conclusions
based upon the identified pattern.
It may be helpful to develop a series of number patterns and have the student identify the pattern and determine
the correct equation. For example, the pattern
5, 7, 9, 11
This sequence follows a pattern of adding 2 to an odd number.
3+2=5
5+2=7
7+2=9
9 + 2 = 11
To practice critical thinking skills, have the student not only find the patterns, but also analyze the
patterns. For instance, have him or her identify the patterns of 3 lists of numbers. Then, ask the student
which pattern will need the fewest additional numbers to reach a specific number. For example, consider
the following 3 patterns. Which pattern will need the fewest additional numbers to reach 95?
(a) 71, 74, 78, 81, 85, ...
(b) 2, 5, 11, 23, 47, ...
(c) 5, 10, 15, 20, 25, ...
For (a), the pattern is to add three and then to add four. It will take 3 more numbers to reach 95. For (b),
the pattern is to double the number and then add 1. It will take 1 number to reach 95. For (c), the pattern
is to add five. It will take 14 numbers to reach 95. Therefore, (a) would be the correct answer.
Example: Which of the following equations can be used to find the numbers in this table?
A. B = A + 9
B. B = 2A + 1
C. B = 3A - 6
D. B = A + 11
Solution: To get from 7 to 15, you must either add 8 or multiply 7 by 2 then add one. To get from 8 to
17, you must either add 9 or multiply 8 by 2 then add one. Since the pattern "multiply by 2 then add
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one" is common in both of the first two rows, check to see if this pattern works for the rest of the rows in
the table. It does work for the rest of the table, so the answer is B.
Functions: Quadratic
A quadratic function is a function that contains polynomial expressions for which the highest power of the
unknown variable is two.
Quadratic functions are written in the form:
or
f(x) is read "f of x."
Here are a few examples of quadratic functions:
Since the value of f(x) (or y) depends on the value of x, the dependent variable is f(x) and the
independent variable is x.
Domain and Range:
The domain of a function is the set of values that can be substituted for the independent variable (x) of a
function. When dealing with the coordinate pairs of a function, the domain is the set of all x values of
the function.
Example 1: Find the domain of the following function.
Since the domain deals with all of the x values of a function, we need to determine all of the x values of
this function. The domain of the function is:
The domain of a function is generally written in numerical order from smallest to largest.
Example 2: Find the domain of
Since any real number that is substituted in place of the independent variable (x) will result in another
real number, the domain of this function is: D = The set of all real numbers.
The range of a function is the set of values that can result from the substitutions for the independent
variable. When dealing with the coordinate pairs of a function, the range is the set of all y values of the
function.
Example 3: Find the range of the following function.
Since the range deals with all of the y-values of a function, we need to determine all of the y values of
this function. The range of the function is
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The range of a function is generally written in numerical order from smallest to largest. If a y-value is
repeated in an ordered pair, it is only listed once in the range.
Example 4: Find the range of
Since any real number that is substituted in place of the independent variable (x) will result in another
real number (g(x)), the range of this function is: R = The set of all real numbers.
Evaluating Quadratic Functions:
In mathematics, evaluate means to substitute values in for the variables and calculate a result. So, to
evaluate a quadratic function, we need to substitute a value in for the independent variable (x) and
calculate the result.
Example 5: Evaluate the quadratic function at f(3).
Step 1: The directions told us to evaluate the quadratic function at f(3), so we substitute 3 in place of all
of the x's in the quadratic function.
Step 2: Following the order of operations, we perform operations on exponents first, then we multiply.
Step 3: Once again, we follow the order of operations which states that multiplying comes before adding
and subtracting.
Step 4: The final step in the order of operations states that adding and subtracting are completed by
working from the left to the right. First we will add 18 and 9. Then we subtract 8 from that sum. We
now know that f(3) = 19.
It is possible to evaluate a function and still have a variable in the answer.
Example 6:
Step 1: Substitute (x + 4) in place of all of the x's in the quadratic function.
Step 2: Applying the order of operations, we must do all operations inside the parentheses. So we add 4
and 3.
Step 3: To take (x + 7) to the second power, we must multiply (x + 7) by itself.
Step 4: In order to multiply (x + 7) by itself, we use the FOIL method. Multiply the "first" terms together
(x times x). Then multiply the "outer" terms (x times 7). Next, multiply the "inner" terms (7 times x).
Finally, multiply the "last" terms (7 times 7).
Step 5: Since 7x and 7x are like terms, we can add them together to get 14x. Also, 49 and 5 are like
terms, so we can add them together to get 54.
Compositions of Quadratic Functions:
The composition of functions is the operation of first applying one function, then applying the other.
Compositions of functions are denoted in two ways. One of those ways is
which can be read, "the
composite of g and f of x." The other way is g(f(x)), which can be read, "g of f of x."
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Example 7:
Step 1: Since we are trying to find "g of f of x," we can substitute
in place of f(x) in g(f(x)) to get
Then, we can substitute that term in place of x in the g(x) = 6x function.
Step 2: Use the distributive property to multiply each term in the parentheses by 6.
Example 8: Find f(g(x)) if f(x) = 2x + 6 and g(x) = 7x - 2.
(1) f(7x - 2) = 2(7x - 2) + 6
(2) f(7x - 2) = 14x - 4 + 6
(3) f(g(x)) = 14x + 2
Step 1: Substitute (7x - 2) in place of g(x) to get f(g(x)) = f(7x - 2). Then substitute (7x - 2) in place of
all x's in the f(x) = 2x + 6 function.
Step 2: Use the distributive property to multiply 7x and -2 by 2.
Step 3: Since -4 and 6 are like terms, they can be added together to make 2. So, f(g(x)) = 14x + 2.
Compositions of functions can be evaluated if we are given the value of the variable.
Example 9:
Step 1: To determine the value of f(g(4)), we must first evaluate g(4). We substitute 4 in place of x
in
Step 2: Following the order of operations we square 4 (4 x 4 =16) and add one to the answer.
g(4) = 17.
Step 3: Now we substitute 17 in place of g(4) to get f(g(4)) = f(17). Then substitute 17 in place of x in
f(x) = 2x - 3.
Step 4: Following the order of operations we multiply 2 and 17 (34), then subtract 3.
Comparing Graphs of Quadratic Functions:
A parabola is the shape of the graph of
The standard form for the equation of a parabola is
(Please remember that f(x) and y can
be substituted for each other in quadratic functions.) When changes are made to an equation such as
adding a constant, changes in the position of the graph of the equation will take place. If the value of a <
0, the graph of the function will reflect itself across the x-axis. The value of h determines whether the
graph will shift right or left and the value of k determines whether the graph will shift upward or
downward. See the chart below.
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Example 10:
(1)
(2) h = -4, k = 7, and a = -1
(3) By looking at the chart, we can determine that the graph of should shift to the left 4 units
(h = -4), upward 7 units (k = 7), and should reflect itself across the x-axis (a = -1).
(4) Compare the actual graphs.
Functions: Absolute Value
The absolute value of a number is the positive distance between the number and 0 on a number line. Absolute
value is denoted by one straight vertical line on each side of a number or expression. Here are some examples of
absolute value notation.
(1) |-6| = 6
(2) |2| = 2
(3) |0| = 0
(4) |x - 3|
(5) |2x + 6|
The absolute value of the first three examples can be determined because they are all integers. Remember the
absolute value of a number is the positive distance between the number and zero on a number line.
An absolute value function is a function which contains at least one absolute value expression. An absolute
value function can be evaluated if the value of the variable is known by substituting the value in for the variable
and calculating the result.
Evaluating Absolute Value Functions:
Example 1: Evaluate y = |x - 7| + 3, for x = 9.
(1) y = |9 - 7| + 3
(2) y = |2| + 3
(3) y = 2 + 3
(4) y = 5
Step 1: Substitute 9 in place of x in the absolute value function.
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Step 2: Before the absolute value can be figured, 7 must be subtracted from 9. This is a rule: All
operations inside an absolute value symbol must be completed before the absolute value can be taken.
Step 3: The distance between 0 and 2 is 2 units, so the absolute value of 2 is 2.
Step 4: The final step is to add 2 and 3 to get 5.
Example 2: Evaluate y = 3|x - 2| + |2x + 4|, for x = -3.
(1) y = 3|(-3) - 2| + |2(-3) + 4|
(2) y = 3|-5| + |-2|
(3) y = 3(5) + 2
(4) y = 15 + 2
(5) y = 17
Step 1: Substitute -3 in place of x in the absolute value function.
Step 2: Compute the values inside both absolute value symbols first.
-3 - 2 = -5
and
2(-3) + 4 = -6 + 4 = -2
Step 3: The distance between 0 and -5 is 5 units, so the absolute value of -5 is 5. The distance between 2 and 0 is 2 units, so the absolute value of -2 is 2.
Step 4: Three times five is 15.
Step 5: Add 15 and 2 to get 17.
Writing an Absolute Value Function as a Compound Function:
Every absolute value function can be written as a compound function. A compound function is a
function that is made up of two or more functions. It is very similar to a compound word in language
arts. In order to write an absolute value function as a compound function, it is necessary to remember
that
An example of this rule would be:
When x = 3, |3| = 3; since
When x = -3, |-3| = -(-3) = 3; since -3 < 0
Example 3: Write the absolute value function, f(x) = |2x + 6|, as a compound function.
Step 1: In order to write an absolute value function as a compound function, it is necessary to determine
the value of x that makes the absolute value function equal to 0. To do this, we set the expression inside
the absolute value symbol equal to zero.
Step 2: Solve 2x + 6 = 0. Subtract 6 from each side of the equal sign to isolate the 2x on one side of the
equal sign.
Step 3: Divide by 2 on each side of the equal sign to isolate the x on one side of the equal sign. If -3 is
substituted into f(x) = |2x + 6|, f(x) = 0.
Step 4:
x would be 2x + 6, so -x would be -(2x + 6) = -2x 6. We use the conditions
because -3 is the value that makes f(x) = |2x + 6| equal zero.
Finding the Vertex of an Absolute Value Function:
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A vertex is a point at which two line segments, lines, or rays meet to form an angle. The graph of every
absolute value function has a vertex. The function f(x) = |x| is graphed below and the vertex is
marked.Remember, the terms f(x) and y can be substituted for each other, so f(x) = |x| can be written as
y = |x|.
The standard form for an absolute value function is y = a|bx + c| + d. It is not necessary to graph an
absolute value function to determine the vertex of the function. If the absolute value function is written
in standard form, we can use the standard form to determine the vertex. The vertex of an absolute value
function is always written as a coordinate point. The x-coordinate of the vertex is where bx + c = 0. The
y-coordinate of the vertex is the value of d.
Example 4: Find the vertex of y = |x - 3| + 2.
(1) x - 3 = 0
(2) x = 3
(3) d = 2
(4) vertex is (3, 2)
Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to
determine the x-coordinate of the vertex.
Step 2: Add 3 to each side of the equal sign to isolate the x and determine that x = 3. The x-coordinate of
the vertex is 3.
Step 3: Since the 2 in y = |x - 3| + 2 is in the same place as the d in the standard form for an absolute
value function, d = 2. The y-coordinate of the vertex is 2.
Step 4: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = |x - 3| + 2 is
(3, 2).
Example 5: Find the vertex of y = -2|4x + 8| - 6.
(1) 4x + 8 = 0
(2) 4x = -8
(3) x = -2
(4) d = -6
(5) vertex is (-2, -6)
Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to
determine the x-coordinate of the vertex.
Step 2: Subtract 8 from each side of the equal sign to isolate the 4x on one side of the equal sign.
Step 3: Divide each side of the equation by 4 to get the x by itself on one side of the equal sign. Since
x = -2, the x-coordinate of the vertex is -2.
Step 4: Since the -6 in y = -2|4x + 8| - 6 is in the same place as the d in the standard form for an absolute
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value function, d = -6. The y-coordinate of the vertex is -6.
Step 5: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = -2|4x + 8| - 6 is
(-2, -6).
Comparing Graphs of Absolute Value Functions:
The standard form for an absolute value function (y = a|bx + c| + d) is needed to compare graphs of
absolute value functions with the graph of y = |x|. The variable a in the standard form determines
whether the graph opens up or opens down. The variables b and c determine whether the graph shifts to
the right or the left and the variable d determines whether the graph shifts upward or downward. See the
chart below.
Example 6: Compare the graph of y = |x| with the graph of y = -|2x + 8| - 3.
(1) In y = -|2x + 8| - 3, a = -1, b = 2, c = 8, and d = -3. Refer back to the standard form of an absolute
value function to see the placements of a, b, c, and d.
(2) By looking at the chart above, we can see that the graph should open downward (a = -1, which is less
than 0). To determine whether the graph shifts left or right, we need to determine c ÷ b = 8 ÷ 2 = 4.
Since 4 is greater than zero, the graph will shift to the left 4 units. The last piece of information we have
is that d = -3. The chart above states that if d < 0, the graph will shift |d| units downward, so this graph
will shift 3 units downward since |-3| = 3.
(3) Compare the actual graphs of y = |x| and y = -|2x + 8| -3.
Matching an Absolute Value Function (Equation) With its Graph:
It is not necessary to graph an absolute value function to match it with its graph. All that is needed is the
standard form for an absolute value function: y = a|bx + c| + d.
Example 7: Match the equation y = -|3x - 9| + 2 with its graph below.
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(1) The first step in matching the equation of an absolute value function with its graph is to determine
the vertex of the graph. The vertex of this graph would be (3, 2) (refer to Example 4 and Example 5 for
an explanation about determining the vertex of an absolute value function).
(2) Now that the vertex is determined, we can eliminate any graph that does not have (3, 2) as its vertex.
Graphs A and D are eliminated.
(3) The next step in matching the equation to its graph is to look at the value of a in the equation.
In y = -|3x - 9| + 2, a = -1. We learned earlier that if a < 0, the graph would reflect itself over the x-axis
(in other words, it will open downward).
(4) Since a = -1 in the equation we are discussing, we can eliminate any graph that does not open
downward. Graph B is not the right graph (D would also be out now if it had not already been
disqualified in step 2).
(5) The only graph left is graph C and it is the correct answer. If there were still two (or more) graphs
left, it would be necessary to substitute some values in for x to determine the value of y and see which of
the graphs contained the calculated points.
Binomial Expansion
A polynomial is the sum of one or more monomials. A monomial is an algebraic expression that has exactly one
term. A binomial is a polynomial that contains exactly two terms. An example of a binomial is (a + b). In this
example, a is considered one term and b is considered another term. Binomial expansion is the result of
rewriting the power of a binomial as a polynomial.
One example of binomial expansion is
There are two ways to expand binomials. Those two methods are: using the Binomial Theorem and
using Pascal's Triangle.
Binomial Expansion Using the Binomial Theorem:
• a and b represent the individual terms of the binomial (such as x and 4)
• n represents the exponent on the original binomial
• r represents the exponent on the "b" term
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•
The other formula that is needed to use the Binomial Theorem is:
In this formula, the n and r represent the same pieces of information they represented in the Binomial
Theorem. The "!" represents a factorial. An example of a factorial is 5! = 5 x 4 x 3 x 2 x 1 = 120.
Example 1: Use the Binomial Theorem to expand the binomial.
First, fill in powers of a and b. The powers of a decrease while the powers of b increase as the terms are
written down.
Second, put in the coefficients. A coefficient is the number that is multiplied by the variables in a term.
For example, 2x + 3y, 2 is the coefficient of x and 3 is the coefficient of y.
There are a few helpful hints for this step in the process. First, in the computation of the coefficients, the
n value always stays the same, but the r value will increase by one as you read from the left to the right.
Second, the exponent on the b term increases by one as you read from the left to the right and ends when
it reaches the value of n. Third, the exponent on the a term decreases by one as you read from the left to
the right and on the final term of the expansion, the exponent of a is zero.
Finally, evaluate the coefficients using
Step 1:
Step 2: Looking at the numerator of the fraction,
4! = 4 x 3 x 2 x 1. We need to evaluate (4 - 1)! in the denominator of the fraction before we can move
on.
Step 3: The numerator of the fraction remains the same as in Step 2. If we concentrate on the
denominator, 3! = 3 x 2 x 1.
Step 4: There are two ways to evaluate
• One way is to multiply to get one number (24) in the
numerator and one number (6) in the denominator. Then divide the numerator (24) by the denominator
(6) to get 4.
• The other way to evaluate the equation is to "cancel out" any numbers that are the same in the
numerator and the denominator.
We will still get 4 as the coefficient.
Continue determining the coefficients for all terms in the binomial expansion in the same manner.
Answer:
Example 2: Use the Binomial Theorem to expand the binomial.
First, fill in powers of a and b, using a = 5x and b = -2y.
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Second, put in the coefficients.
Third, evaluate the coefficients using
Fourth, evaluate all exponents. Remember that
Finally, multiply to determine the binomial expansion of
Answer:
Binomial Expansion Using Pascal's Triangle:
Pascal's triangle can be thought of as a two-dimensional sequence. Each term is determined by a row and
its position in that row. The only term in the top row (row 0) is 1. The first and last elements in all other
rows are also 1. If x and y are located next to each other on a row, the element just below and directly in
between them is x + y. See the example below.
Pascal's triangle helps us with binomial expansion by giving us the coefficients of each term of the
binomial expansion. The number of the row of Pascal's triangle corresponds to the exponent of the
binomial and the entries in that row of the triangle are the coefficients of the terms in the binomial
expansion. The table below illustrates how Pascal's triangle corresponds to binomial expansions.
Example 3: Use Pascal's Triangle and the Binomial Theorem to expand the binomial.
First, fill in the powers of a and b, using a = 2x and b = 4.
Second, put in the coefficients.
Third, use Pascal's Triangle to determine the coefficients. Since the original binomial that we want to
expand has an exponent of 4, we want to look at the 4th row of Pascal's Triangle for the values of the
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coefficients. The elements in the fourth row of Pascal's Triangle are: 1
4
6
4
1, so
Fourth, evaluate all exponents.
Fifth, multiply to determine the binomial expansion of
Answer:
Finding the Coefficient of One Term in a Binomial Expansion:
It is possible to use the Binomial Theorem and Pascal's Triangle to determine the coefficient of a single
term in a binomial expansion.
Example 4:
First, we only need the coefficient of one term, so we do not need to fill in the entire expansion. We
only need the portion of the expansion where the x has an exponent of 5. We can determine which term
we need to focus on by remembering that the sum of the exponents of each term of this binomial should
equal 7 (the exponent of the binomial we are expanding). The exponent of the 6 must be 2 (5 + 2 = 7 or
7 - 2 = 5), since we want the
The portion of the expansion we are dealing with is:
Second, we need to determine the coefficient of the term with
That coefficient would be
now know that the portion of the expansion we are dealing with is:
We
Third,
We need to find the 3rd element (reading from left to
right) of the 7th row of Pascal's Triangle because
would be the third term of the binomial expansion. The third element of the
seventh row of the triangle is 21. The portion of the expansion should now be:
Finally, we must evaluate the powers and multiply to determine the coefficient of the
is 36. Thirty-six times 21 is 756. The term is now
Answer: 756
Six squared
Example 5:
We refer back to Pascal's Triangle.
The variable n represents the row in Pascal's Triangle we need to look at. The expression
(r + 1) represents the term (reading from left to right) we need to find in row n.
So, to determine the binomial coefficient, we look at the 6th row, 4th term of Pascal's Triangle. That
term is 20.
Answer: 20
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Non-Linear Equations
A non-linear equation is an equation whose graph is not a straight line.
An example of a linear equation (graph is a straight line) is y = x + 3 and an example of a non-linear
equation is
These two equations are graphed below.
A basic rule to follow to determine whether an equation is linear or non-linear is that non-linear
equations have variables that have powers other than one and linear equations have powers equal only to
one.
Simplifying Square Roots:
Before we can begin solving non-linear equations, we must discuss how to simplify square roots. When
a number is multiplied by itself the product is the square of the number. A square root of a number is a
factor that when multiplied by itself equals the number. For example: 2 x 2 = 4, so 4 is the square of 2
also, since 4 = 2 x 2, 2 is a square root of 4. Another square root of 4 is -2 because -2 x -2 = 4. Numbers
that have a rational number as their square root are called perfect squares. Examples of perfect squares
are: 9 (square root is 3), 16 (square root is 4), 25 (square root is 5) and 9/25 (square root is 3/5).
The notation for a square root is this symbol:
To simplify a square root, we first determine two factors that multiply to make the whole number. One
of these two factors should be a perfect square, preferably the largest perfect square that is a factor of the
number. Then we take the square root of the perfect square factor and place that number in front of the
radical symbol.
Example 1: Simplify.
Step 1: Rewrite the problem as the square root of two factors of 72. Remember, one of the factors
should be the largest perfect square that is a factor of 72. In this case 36 and 2 were used because 36
times 2 equals 72 and 36 is a perfect square.
Step 2: Now the problem can be rewritten as the square root of 36 times the square root of 2.
Step 3: Determine the square root of 36 (which is 6) and multiply it by radical 2. Since 2 does not have a
factor that is a perfect square, radical 2 does not change.
Example 2: Another way to simplify
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Step 1: It is possible to simplify a square root if the largest perfect square factor is not known. Once
again, rewrite the problem as the square root of two factors of 72. Make sure one of these two factors is
a perfect square. In this case 9 and 8 were used because 9 is a perfect square and 9 times 8 equals 72.
Step 2: Since 8 has a factor that is a perfect square (4), the problem must be rewritten as the product of
these three factors of 72. 9 x 4 x 2 = 72.
Step 3: Since 9 and 4 are perfect squares and 2 does not have a perfect square factor, the problem can be
rewritten as the square root of 9 times the square root of 4 times the square root of 2.
Step 4: Determine the square root of 9 (which is 3), multiply it by the square root of 4 (which is 2), and
multiply them both by radical 2.
Step 5: Finally, multiply 3 and 2 to get 6. The 6 is multiplied by radical 2 to obtain the final answer:
Solving Non-Linear Equations:
Non-Linear equations can be solved in much the same way as linear equations. The goal of solving a
non-linear equation is to isolate the variable on one side of the equal sign.
Example 3: Solve the following equation.
Answer:
Example 4: Solve the following equation.
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Using the Pythagorean Theorem to Solve Non-Linear Equations:
The Pythagorean Theorem can be used to determine the length of a missing side of a right triangle. A
right triangle is a triangle that has one right (90º ) angle. A 90º angle is marked in a triangle with a box
in the angle. The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle is
equal to the square of the hypotenuse. The hypotenuse of a right triangle is the side of the triangle
opposite the right angle. The other two sides of the triangle are called the legs.
The Pythagorean Theorem states:
In the Pythagorean Theorem, 'c' represents the length of the hypotenuse and 'a' and 'b' represent the
lengths of the legs of the right triangle. The Pythagorean Theorem only works for right triangles.
Example 5: Use the Pythagorean Theorem to solve for x.
Step 1: Write the Pythagorean Theorem. Then determine the values of a, b, and c. Remember, c always
represents the length of the hypotenuse. In this triangle, the length of the hypotenuse is 8, so c = 8. It
does not matter whether a or b is assigned the value of x or the value of 6.
Step 2: Substitute the values of a, b, and c into the Pythagorean Theorem.
Step 3: Following the order of operations, square the 8 (8 x 8 = 64) and the 6 (6 x 6 = 36).
Step 4: Subtract 36 from each side of the equation.
Step 5: Take the square root of each side of the equation.
Step 6: Simplify radical 28. In this case, the variable represents the length of the side of a triangle, so the
answer cannot be negative.
Answer:
Solving an Equation by Completing the Square:
Completing the square is a method of solving a quadratic equation in order to express the equation as a
single squared term. The method of completing the square is used when an equation cannot be factored.
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Completing the square involves adding the square of one term to the equation and solving the equation
for the value of the variable. The theorem for completing the square states:
The following is a detailed example of how to complete the square.
Example 6: Solve the equation by completing the square.
Step 1: Add 7 to each side of the equation to put the equation in the form
Step 2: Since 8 is in the
same place as b, b = 8. Add the square of 1/2 of b to each side of the equation. One-half of 8 equals 4, so
we are actually adding 4 squared to each side of the equation.
Step 3: We need to fill in the final form in the theorem for completing the square. One-half of b equals
4, so inside the parentheses we have (x + 4). Then we place an exponent of 2 outside the parentheses.
Finally, 4 squared equals 16 (4 x 4).
Step 4: First, add 7 and 16 to get 23. Then take the square root of each side of the equation. Taking the
square root of a term is the opposite of squaring a term, so we get x + 4 on one side of the equal sign.
Remember, every number has a positive and a negative square root, so we write
Step 5: Subtract 4
from each side of the equation to isolate the x on one side of the equal sign. The new expression cannot
be simplified.
Step 6: The solution to the equation is
Characteristics that Describe the Graph of a Non-Linear Equation:
A quadratic equation is any equation in the form
The graph of a quadratic equation is always a
parabola. The vertex of a parabola can be found by putting the quadratic equation in vertex form.
Once a quadratic equation is in vertex form, the vertex is the coordinate point (h, k). If a > 0, then the
graph opens up and has a minimum. If a < 0, then the graph opens down and has a maximum.
The roots of a quadratic equation are the solutions of the quadratic equation when y = 0. The roots are
the points where the graph intersects the x-axis; therefore, y = 0. The axis of symmetry is the line that
passes through the vertex and splits the graph directly in half such that each side is the mirror image of
the other. This line is represented by the equation x = h.
Example 7: What are the characteristics of the graph of
Step 1: Rewrite the equation in vertex form. The simplest way to do this is by completing the square.
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Step 1A: Subtract 4 from each side of the equation.
Step 1B: Factor -1 out of the right side of the equation to make the term positive.
Step 1C: Add 1/2 of the b term (-6) squared to the right side of the equation and subtract 1/2 of the b
term from the left side of the equation (we subtract on the left side of the equation because we factored 1 out of the equation in Step 1B and we need to multiply any number by -1 before we can add or
subtract it from the left side of the equation). b = -6, so we are actually adding -3 squared to the right
side of the equation and subtracting -3 squared from the left side of the equation.
Step 1D: Complete the theorem for completing the square.
Step 2: Determine whether the graph opens up or down. Now that the equation is in vertex form, we can
determine that a = -1. Since a < 0, the graph opens down.
Step 3: Determine the vertex of the parabola. The vertex is the point (h, k). h = 3 and k = 13, so the
vertex is (3, 13).
Step 4: Determine the axis of symmetry. The axis of symmetry is the line x = h. The axis of symmetry is
x = 3.
Step 5: Determine the roots of the equation. The roots of the equation are the values of x when y = 0.
Step 5A: Substitute 0 in place of y.
Step 5B: Divide each side of the equation by -1.
Step 5C: Take the square root of each side of the equation.
Step 5D: The roots of the equation are
If the 13 had been negative, the equation would have had no real roots.
(1) opens down
(2) vertex is (3, 13)
(3) axis of symmetry is: x = 3
(4) roots are
Graphing Functions
A function is a relation between two variables such that each value of the first variable corresponds to
exactly one value of the second variable. A polynomial function is a function that involves a series of
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terms added together. A polynomial function can be expressed in standard form as
where n represents the degree of the polynomial. (Remember, when you are working with functions f(x)
and y can be interchanged.) The degree of a function is the largest exponent on the variable.
Graphs of Polynomial Functions:
The graph of a polynomial function has at most (n - 1) turns, that is, it can change direction at most (n 1) times. The direction a graph is going (up or down) is always read from the left to the right. Graph A
in the diagram below changes direction once; whereas, graph B changes direction three times.
The graph of a continuous function will have no breaks. The graphs in the diagram above are graphs of
continuous functions. A graph of a function that is not continuous is shown below.
All polynomial functions are continuous functions.
Example 1: State the maximum number of turns in the graph of the polynomial.
(1) n = 5
(2) 5 - 1 =4
Step 1: The degree of the polynomial is 5. The highest exponent on the x is 5.
Step 2: n - 1 = 4 so the maximum number of turns is 4.
Vertical and Horizontal Asymptotes of Graphs:
An asymptote is a line that the graph of a function comes close to, but never touches. For a rational
function (a function with a numerator and a denominator), f(x) = g(x)/h(x), where g(x) and h(x) have no
common factors and h(x) does not equal zero. It is possible to have vertical and/or horizontal asymptotes
in rational functions. In the diagram below, the line y = 1 is a horizontal asymptote and the lines x = -3
and x = 3 are vertical asymptotes. Asymptotes are represented by dotted lines because points on the
asymptotes are not points that lie on the graph of the function.
The vertical asymptotes are the zero(s) of h(x).
The horizontal asymptotes are as follows:
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• The line y = 0 is the horizontal asymptote if the degree of g(x) < degree of h(x).
• The line y = a/b is the horizontal asymptote, where a is the leading coefficient of g(x) and b is the
leading coefficient of h(x) if the degree of g(x) = degree of h(x).
• There is no horizontal asymptote if the degree of g(x) > degree of h(x).
A coefficient is the number that is multiplied by the variable(s) in a term. The leading coefficient of a
polynomial is the coefficient of the first term when the polynomial is written in standard form
(remember, the standard form for a polynomial has the exponents in descending order).
Example 2:
Find the vertical asymptotes first. Note that
(1) x - 1 = 0
(2) x = 1
Step 1: Solve for the zeros of h(x) by setting the denominator equal to zero.
Step 2: Add 1 to both sides of the equation to solve for x. The vertical asymptote is at x = 1.
Find the horizontal asymptotes.
(1) The degree of the numerator is 2 and the degree of the denominator is 1.
(2) Since the degree of the numerator is greater than the degree of the denominator, there is no
horizontal asymptote.
The function f(x) has only one asymptote. It is a vertical asymptote at x = 1.
Example 3:
Find the vertical asymptotes first. Note that
Step 1: Solve for the zeros by setting the denominator equal to zero.
Step 2: Add 4 to both sides of the equation to solve for x.
Step 3: Take the square root of both sides of the equation to solve for x. Since 4 has two square roots, 2
and -2, there are two vertical asymptotes, x = 2 and x = -2.
Find the horizontal asymptotes.
(1) The degree of the numerator is 2 and the degree of the denominator is 2.
(2) The degree of the numerator is equal to the degree of the denominator. The leading coefficient of the
numerator is 3 (so, a = 3) and the leading coefficient of the denominator is 1 (so, b = 1). Therefore, the
horizontal asymptote is the line y = a/b = 3/1 or y = 3.
Answer: The function f(x) has three asymptotes: x = 2, x = -2, and y = 3.
Example 4:
Find the vertical asymptotes first.
(1) x - 3 = 0
(2) x = 3
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Step 1: Solve for the zeros by setting the denominator equal to zero.
Step 2: Add 3 to both sides of the equation to solve for x. The vertical asymptote is x = 3.
Find the horizontal asymptotes.
(1) The degree of the numerator is 0 (because the numerator is a constant) and the degree of the
denominator is 1.
(2) Since the degree of the numerator is less than the degree of the denominator, the horizontal
asymptote is the line y = 0.
Answer: The function f(x) has two asymptotes: x = 3 and y = 0.
Right and Left Behaviors of a Graph:
There are four basic rules for determining the right and left hand behaviors of a graph of a function.
Those rules are listed below.
•
•
•
•
If the leading coefficient of a polynomial function is positive then the graph rises to the right.
If the leading coefficient of a polynomial function is negative then the graph rises to the left.
If the degree of the function is even, then the graph has the same right and left behavior.
If the degree of the function is odd, then the graph has opposite right and left behaviors.
Example 5: Describe the left and right behaviors of the graph of
The leading coefficient is positive, and the degree is 3, an odd number. The graph will rise to the right
and fall to the left.
Example 6: Describe the left and right behaviors of the graph of
(1) The leading coefficient is positive, so the graph will rise to the right.
(2) The degree is 2 (an even number), so the graph has the same right and left behavior. This means that
the graph will also rise to the left.
The graph will rise to the right and rise to the left.
Horizontal and Vertical Shifts of Graphs:
When certain changes to an equation of a function are made, the graph of the new function may vary by
moving the original graph to a new location in the coordinate plane, while the basic shape of the graph
remains the same. The horizontal and vertical shifts of a graph caused by changes made to an equation
can be summarized as follows.
If f(x) is a function:
• the graph of f(x) + a will result in a vertical shift upward of |a| units
• the graph of f(x) - a will result in a vertical shift downward of |a| units
• the graph of f(x + b) will result in a horizontal shift to the left |b| units
• the graph of f(x - b) will result in a horizontal shift to the right |b| units
• the graph of -f(x) will result in a reflection in the x-axis
A change to an equation can result in both a horizontal and vertical shift such as the graph of f(x + b) + a
which will result in a vertical shift up |a| units and a horizontal shift to the left |b| units.
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Example 7:
Since the shift follows the f(x) + a formula, where a = 9, the graph will shift 9 units up.
Example 8:
Since the shift follows the f(x + b) - a formula, where a = -4 and b = 3, the graph will shift 4 units down
and 3 units to the left.
Example 9:
Since the shift follows the -f(x - b) + a formula, where a = 1 and b = -2, the graph will reflect across the
x-axis, shift 1 unit up and 2 units to the right.
Example 10: Compare the graph of g(x) = |x| - 3 to the graph of f(x) = |x|.
Since the shift follows the f(x) - a formula, where a = -3, the graph will shift 3 units down.
Example 11: Compare the graph of g(x) = |x + 1| + 4 to the graph of f(x) = |x|.
Since the shift follows the f(x + b) + a formula, where a = 4 and b = 1, the graph will shift 4 units up and
1 unit to the left.
Rational Expressions: Add/Subtract
A rational expression is a fraction whose numerator and denominator are polynomials. To add or subtract a
rational expression, the denominators must be the same. If the denominators are the same, add or subtract the
numerators and keep the common denominator.
Example 1: Subtract the fractions.
Step 1: Rewrite the expression as one fraction. The numerators will be subtracted and the denominator
will remain the same.
Step 2: Subtract the numerators. 3 - 5 = -2
Step 3: Since 2 and 8 can both be divided by 2, the fraction can be reduced to -1/4x.
Example 2: Add the fractions.
Step 1: Rewrite the expression as one fraction. The numerators will be added and the denominator will
remain the same.
Step 2: Add the like terms, 2x and 5x, to get 7x. It is not necessary to make any changes to the
denominator.
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Example 3:
Step 1: Rewrite the expression as one fraction. The denominator is the same on all three fractions, so it
will remain the same.
Step 2: Collect the like terms together.
Step 3: Add 8t and 4t together to get 12t, and -14 and -8 together to get -22. It is not necessary to make
any changes to the denominator.
To add or subtract rational expressions with unlike denominators, find the lowest common denominator
of the expressions. Then rewrite each expression as an equivalent rational expression using the lowest
common denominator. Once the denominators are like, add or subtract the numerators and keep the
common denominator.
To find the lowest common denominator, find the least common multiple of the denominators. Begin by
factoring each denominator, if possible. Then multiply the individual factors together to determine the
lowest common denominator of the fractions.
Example 4: Add the fractions.
Step 1: Determine the common denominator. To do this, list the denominator of each fraction, and break
them down into their factors.
Step 2: Multiply the factors that the denominators have in common: 3 and We now need to multiply
by 6 because it is the only factor left that has not been multiplied. The lowest common denominator is
Step 3: Rewrite the fractions with the common denominator. The fraction on the left already has the
lowest common denominator as its denominator, so we do not need to change it, but we do need to
change the fraction on the right. We multiply the numerator and denominator by 6 to get
Step 4: Use
the distributive property to multiply (5x + 2) by 6. This involves multiplying 5x by 6 to get 30x and
multiplying 2 by 6 to get 12.
Step 5: Rewrite the fractions as one fraction. Since there are no like terms in the numerator of the
fraction, the numerator remains as it is and the denominator does not change.
Example 5: Add the fractions.
Step 1: Determine the common denominator. Since the whole number 7 has a denominator of 1, the
common denominator of these two fractions is
Rewrite 7/1 as a fraction with the common
denominator. This involves multiplying the numerator and denominator of the fraction by
Step 2:
Multiply to make the fractions have common denominators.
Step 3: Add the numerators.
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Step 4: Since the two terms in the numerator can both be divided by x-squared, factor x-squared out of
the terms.
Step 5: Simplify the fraction by cancelling out the x-squared in the numerator and two of the x's from
the x to the fifth power in the denominator.
Example 6: Subtract the fractions.
Step 1: Factor the denominator of the second fraction. The factors of the denominator are (m + 3) and (m
+ 2).
Step 2: Rewrite the fractions with the second denominator in factored form.
Step 3: Determine the common denominator. Since each denominator has (m + 3) as a factor, the
common denominator is (m + 3)(m + 2). Multiply the numerator and denominator of the first fraction by
(m + 2) to rewrite the fraction with the common denominator. The second fraction does not need to be
changed because it already has the common denominator.
Step 4: Distribute m to (m + 2). This involves multiplying each term in (m + 2) by m. Now the fractions
have common denominators.
Step 5: Rewrite the fractions as one fraction. Remember to place the numerator of the second fraction in
parentheses because the entire numerator is being subtracted from the numerator of the first fraction.
Step 6: Distribute the subtraction symbol to each term in the parentheses. This involves changing the
sign of each term in the parentheses.
Step 7: Collect the like terms in the numerator.
Step 8:
Example 7: Add the fractions.
Step 1: Factor each denominator. Each term in the denominator of the first fraction can be divided by 5,
so 5 is factored out of the expression. Each term in the denominator of the second fraction can be
divided by 3x, so 3x is factored out of the expression.
Step 2: Rewrite the fractions with the denominators factored.
Step 3: Each denominator has a factor of (x - 4), so that is automatically part of the denominator. The
first fraction also has a factor of 5, so the numerator and denominator of the second fraction must be
multiplied by 5 to acquire the common denominator. The denominator of the second fraction also has a
factor of 3x, so the numerator and denominator of the first fraction must be multiplied by 3x to acquire
the common denominator. The common denominator of these two fractions is 15x(x - 4).
Step 4: Perform the necessary multiplications. Now the fractions have common denominators and can be
added together.
Step 5: Rewrite the fractions as one fraction. The numerators are added together. There are no like terms
in the numerator, so the numerator cannot be simplified.
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Example 8: Subtract the fractions.
Step 1: Factor the denominators. Since the denominator of the second fraction (4x - 5) cannot be
factored, it is left alone. The denominator of the first fraction can be factored using the "difference of
squares" rule. The factors are (4x - 5) and (4x + 5).
Step 2: Rewrite the fractions with the factored denominators.
Step 3: Determine the common denominator. Each fraction has (4x - 5) as a factor in the denominator,
so (4x - 5) is part of the common denominator. The denominator of the first fraction also has (4x + 5) as
a factor, so the numerator and denominator of the second fraction need to be multiplied by (4x + 5) to
acquire the common denominator. The common denominator is (4x - 5)(4x + 5).
Step 4: Perform the necessary multiplications. Now the fractions have common denominators and can be
subtracted.
Step 5: Rewrite the fractions as one fraction. Remember to place the numerator of the second fraction in
parentheses because the entire numerator is to be subtracted from 6x.
Step 6: Distribute the subtraction symbol through the parentheses. This involves changing the sign of
each number in the parentheses.
Step 7: Collect the like terms together and simplify. Since 6x and -36x are like terms, they can be added
together to get -30x.
Example 9: Add the fractions.
Step 1: Factor the denominators. Then use the factors to determine the common denominator. The
common denominator is x(x - 1)(x - 2).
Step 2: Multiply the numerator and denominator of the first fraction by x and the numerator and
denominator of the second fraction by (x - 1).
Step 3: Perform the necessary multiplications. Now the fractions have common denominators and can be
added.
Step 4: Rewrite the fractions as one fraction.
Step 5: Collect like terms.
Step 6: Since both terms in the numerator can be divided by 2x, we can factor 2x out of the numerator.
Step 7: The x (from 2x) in the numerator of the fraction and the x in the denominator of the fraction
divide out to simplify the fraction.
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Rational Functions: Multiply/Divide
A rational expression is a fraction whose numerator and denominator are polynomials. A rational expression is
in its simplest form when the numerator and denominator have no common factors.
Simplifying Fractions:
If a, b, and c are nonzero real numbers, then
You can only divide out factors, not terms that are being added or subtracted. For example,
cannot
be simplified because the x in the numerator is being added to the 1, so it is not a factor. Since the x in
the numerator is not a factor, it cannot divide out with the x in the denominator.
To simplify a rational expression, factor the numerator and denominator. Then divide out any common
factors.
Factoring:
There are four types of factoring. These types are listed below with an example of each type.
1. Factoring out the Greatest Common Monomial Factor:
2. Factoring the Difference of Squares:
This factorization works because:
3x times 3x equals
-12x plus 12x equals 0x.
-4 times 4 equals -16
3. Factoring a Perfect Square Trinomial:
This factorization works because:
x times x equals
10x plus 10x equals 20x.
10 times 10 equals 100
4. Factoring a Trinomial:
This factorization works because:
x times x equals
-7x plus -2x equals -9x.
-7 times -2 equals +14
Remember to always factor out the greatest common monomial factor first, if there is one.
Example 1:Factor the following polynomial.
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Step 1: The three terms in the polynomial can be divided by 3, so we need to factor 3 out of the
polynomial. Then write the new polynomial.
Step 2: Factor the polynomial that remains in the parentheses. The factors of the polynomial in
parentheses are (x + 2) and (x + 10). The polynomial is now completely factored.
Answer: 3(x + 2)(x + 10)
Example 2: Simplify.
Step 1: Factor the numerator and the denominator. The three terms in the numerator do not have a
common factor. The numerator must be factored using the factoring a trinomial method. The two terms
in the denominator have a common factor of x. Divide x out of both terms to factor the denominator
completely.
Step 2: Rewrite the expression with the numerators and denominators in factored form.
Step 3: The (x - 3) in the numerator and the (x - 3) in the denominator divide out to simplify the
expression. The expression is now completely simplified.
Answer: (x - 2)/x
Multiplying Rational Functions:
To multiply rational expressions, multiply the numerators, multiply the denominators, and then simplify.
Example 3: Multiply and simplify.
Step 1: Rewrite the fractions as one fraction. The numerators should be multiplied and the denominators
should be multiplied.
Step 2: Multiply the numerators and the denominators. Remember that when multiplying numbers taken
to powers that have the same bases, the exponents are added, so
and
Step 3:
Divide out common factors. 20 and 84 can both be divided by 4. The in the numerator and
denominator divide out. When two terms with the same base are divided, their exponents are subtracted,
so
Example 4: Multiply and simplify.
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Step 1: Factor the numerators and denominators, if possible.
is a perfect square trinomial and has
factors of (x + 3)(x + 3). Each of the terms of
has a factor of x, so the expression factors as x(x + 1).
Neither the numerator nor the denominator of the second fraction can be factored, so they are left as they
are.
Step 2: Rewrite the fractions with the factors in the correct places.
Step 3: Rewrite the fractions as one fraction. Remember to place the numerator and denominator of the
second fraction in parentheses since they are to be multiplied.
Step 4: Divide out common factors. One of the (x + 3) terms in the numerator and the (x + 3) term in the
denominator divide out. The (x + 1) term in the numerator divides out with the (x + 1) term in the
denominator. The fraction is now completely simplified.
Example 5: Multiply and simplify.
Step 1: Rewrite the term (x - 7) as a fraction with 1 as the denominator and enclose all terms in
parentheses.
Step 2: Multiply the numerators and the denominators.
Step 3: Divide out common factors.
Dividing Rational Expressions:
To divide one rational expression by another, multiply the first expression by the reciprocal of the
second expression.
The reciprocal of a non-zero number
Example 6: Divide and simplify.
Step 1: Rewrite the problem by multiplying the first expression by the reciprocal of the second
expression.
Step 2: Factor the numerators and denominators, if possible. Neither of the numerators can be factored.
The denominator of the first expression is a perfect square trinomial that factors as (x + 2)(x + 2). The
denominator of the second expression is a factorable trinomial that factors as (x + 1)(x + 7).
Step 3: Multiply the numerators and the denominators. Remember to place the numerators in
parentheses because they are complete terms.
Step 4: Divide out common factors. When all factors of the numerator divide out, the numerator is 1.
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Example 7: Divide and simplify.
Step 1: Rewrite the first expression as a fraction with 1 as the denominator.
Step 2: Rewrite the problem by multiplying the first expression by the reciprocal of the second.
Step 3: Factor the numerators and denominators, if possible.
Step 4: Multiply the numerators and the denominators. Remember to place the 12x in parentheses.
Step 5: Divide out common factors.
Step 6: Since 12 and 3 can both be divided by 3, we can still simplify. 12 ÷ 3 = 4 and 3 ÷ 3 = 1. Now
the expression is completely simplified.
Multiplication/Division Rational No.
The following numbers are rational numbers because they can all be written as fractions.
Multiplying and dividing rational numbers includes the calculation of whole numbers, fractions,
decimals, and integers. To understand how to multiply and divide rational numbers, the student needs to
know the following rules:
1. A positive rational number multiplied by a positive rational number equals a positive rational number.
(+0.5) x (+0.2) = +0.1
2. A negative rational number multiplied by a positive rational number equals a negative rational
number. A positive rational number multiplied by a negative rational number equals a negative rational
number.
(-0.5) x (+0.2) = -0.1
3. A negative rational number multiplied by a negative rational number equals a positive rational
number.
(-0.5) x (-0.2) = +0.1
4. A positive rational number divided by a positive rational number equals a positive rational number.
(+0.1) ÷ (+0.2) = +0.5
5. A negative rational number divided by a positive rational number equals a negative rational number.
A positive rational number divided by a negative rational number equals a negative rational number.
(-0.1) ÷ (+0.2) = -0.5
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6. A negative rational number divided by a negative rational number equals a positive rational number.
(-0.1) ÷ (-0.2) = +0.5
Rational Numbers: Equations
Rational numbers are numbers that can be written as fractions. The following numbers are rational
numbers because they can all be written as fractions.
Adding and subtracting rational numbers includes the calculation of whole numbers, fractions, decimals,
and integers. It is common for students learning to solve equations with rational numbers to confuse
operational symbols with integer signs.
Example 1: Before solving, n - - 7 = - 9, the equation should be simplified to n + 7 = - 9. Solve for n in
n + 7 = - 9.
Solution: To solve for n, you need to first isolate n on one side of the equal (=) sign by subtracting 7
from both sides.
Answer: n = - 16
Example 2: Solve for x.
Step 1: Multiply both sides of the equation by the reciprocal of -2/3, which is -3/2. This will isolate the x
on one side of the equal sign.
Step 2: Make the -6 into a fraction by giving it a denominator of 1 and multiply the two fractions.
Remember, multiply numerator by numerator (6 x -3 = -18) and denominator by denominator (1 x 2 =2).
Step 3: Reduce -18/2 to lowest terms by dividing -18 and 2 by 2.
Answer: -9
Equations: Addition/Subtraction
Equations are number sentences which contain equal signs.
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified.
Example: y - 6
Variables are letters or symbols that represent numbers that are unknown.
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The following are examples of how to solve equations.
Example 1: Solve: n + 6 = 20
Step 1: Write the equation.
Step 2: Subtract 6 from both sides of the equation.
Answer: n = 14
Example 2: Solve for x.
16.6 = 5.5 + x - 8.2
Step 1: Begin to isolate the variable on one side of the equation by adding 8.2 to both sides.
Step 2: Subtract 5.5 from both sides to completely isolate the variable.
Answer: x = 19.3
Equations: Multiplication/Division
Equations are number sentences which contain equal signs:
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified:
Example: y - 6
A common mistake among students first learning to multiply and divide equations is the failure to use
inverse operations. For instance, in the equation 3x = 8, the student might multiply 3 to both sides of the
equal sign. The correct procedure is to use the inverse operation and divide 3 from both sides (x = 8/3).
By applying the inverse operation, the student can isolate the x.
Example 1: Solve: 5n = 75
Step 1: In the equation, solve for the value of n.
Step 2: 5n is the same as 5 x n. Divide both sides of the equation by 5.
Step 3: 75 ÷ 5 = 15
Answer: n = 15
Example 2: Solve for t.
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Step 1: Multiply both sides of the equation by 16 to eliminate the fraction.
Step 2: Divide both sides if the equation by 4 to isolate the variable t.
Answer: t = 128.
Equations: Writing
Writing equations demonstrates the ability to understand and apply equations in real world situations. To write
equations, students must interpret data presented in word problems and formulate numerical equations.
Equations are number sentences which contain equal signs.
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified.
Example: y - 6
The following is a step-by-step example.
Example: Today, Jim worked 2 minutes longer than he did yesterday. Yesterday, he worked for 19 minutes.
How long did Jim work today?
(1) yesterday (y) = 19 and today (t) = y + 2
(2) t = y + 2
(3) t = 19 + 2
Step 1: Identify the known amounts.
Step 2: Develop an equation from the known amounts.
Step 3: Using substitution, insert the known amounts and solve.
Answer: t = 21 minutes
Story Problems
Some story problems present equation and expression problems in text format.
Equations are number sentences which contain equal signs.
Example: n + 5 = 9
Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified.
Example: y - 6
Example: Avital ran 10 miles more than she did yesterday. Yesterday, she ran 9 miles. How many miles did
Avital run today?
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(1) yesterday (y) = 9 and today (t) = y + 10
(2) t = y + 10
(3) t = 9 + 10 = 19
Step 1: Identify the known amounts.
Step 2: Develop an equation from the known amounts.
Step 3: Using substitution, insert the known amounts and solve.
Answer: Avital ran 19 miles today.
Inequalities - B
An inequality is a number sentence that uses "is greater than", "is less than", or "is not equal to" symbols. For
example, 6n > 4 is a number sentence with an inequality symbol.
It may be useful to review the inequality symbols.
Example 1: Solve for y.
8y > 40
Get the variable being solved for (y) on one side of the inequality and the whole number on the other.
To do this, divide both sides by 8.
The correct answer is that y is greater than 5.
Inequalities can be represented as a value on a number line. The following number line represents the
inequality
Example 2: Which inequality represents the value shown on the number line below?
A. n < 3
B. n > 3
C. n = 3
The answer is A. n < 3 because the dot on the number line is open.
Example 3: Solve the following inequality.
-3 < -2x - 3 < 23
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Step 1: We can solve by separating the inequality into two inequalities as shown below.
-3 < -2x - 3 and -2x - 3 < 23
Step 2: Solve both inequalities.
The answer is -13 < x < 0.
NOTE: -13 < x < 0 is the same as 0 > x > -13.
Equations: Two-Step
Two-step equations require students to perform two operations before solving the equation.
For example, the equation 3x - 3 = 18 requires adding 3 to both sides of the equation and dividing both sides of
the equation by 3.
Example 1: Find the value of n in the equation 6n - 9 = 9
Step 1: In the equation, solve for the value of n by first getting n alone. Add +9 to both sides of the
equation.
Step 2: Divide both sides of the equation by 6.
Answer: n = 3
Example 2: Solve for t.
(1/8)t + 3 = 12
Step 1: To get the variable alone, subtract 3 from both sides of the equation.
Step 2: Isolate the 't' by multiplying by the reciprocal of 1/8, which is 8/1.
Answer: t = 72
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Solving Equations: Substitution
Letters expressing unknown values in equations are called variables. In these two variable equations, students
are given the value of one variable. To find the value of the second variable, substitution must be used.
If y = 4x, and x = 9, substitute the given value of x, in the equation, y = 4x. The result is y = 4(9). Calculate the
right side of the equation to get y = 36. Now the values of both variables are known: x = 9 and y = 36.
To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each side of
the equation by 4. The result is x = 9, so the values for x and y are both correct.
Example 1: Solve for x.
x = 3y + 9
y = -2
Solution: Substitute -2 in place of 'y' in the equation. Then, solve for x.
x = 3(-2) + 9
x = -6 + 9
x=3
When we are not given enough information, the problem cannot be solved.
Example 2: Solve for t.
t = s -12
r=s+3
Solution: Since we do not know the value of either 's' or 'r', we cannot determine the value of t.
Rates
Rates refer to the price per unit of a particular commodity. Pricing tactics often cause confusion for today's
consumer. Products are sold at 3 for one amount or 5 for another. Calculating rates allows students to figure out
the exact cost of the items.
Example: Apples are 5 for $1.25 and bananas are 6 for $1.80, which fruit is less expensive if you only buy one?
(1) $1.25 ÷ 5 = $0.25 --> cost of one apple
(2) $1.80 ÷ 6 = $0.30 --> cost of one banana
(3) $0.25 < $0.30
Step 1: Determine the cost of one apple by dividing 1.25 by 5. One apple costs $0.25.
Step 2: Determine the cost of one banana by dividing 1.80 by 6. One banana costs $0.30.
Step 3: Compare the cost of the two pieces of fruit. $0.25 is less than $0.30.
Answer: One apple is less expensive than one banana.
Multiple-step Story Problems - F
These problems are designed to assess a student's ability to interpret data from word problems (some contain
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both relevant and irrelevant data) and develop solutions which require two or more operations (or steps) to
solve.
It may be helpful to develop a series of multiple step word problems that relate to the student's activities, such
as allowance. The following is a step-by-step example of a multiple-step story problem.
Example 1: Shinika bought a dozen eggs and 3 gallons of milk at the grocery store. If each gallon of milk was
$2.60 and the total bill was $10.25, how much is one egg?
(1) $10.25 - 3($2.60) = ?
(2) $10.25 - $7.80 = ?
(3) $10.25 - $7.80 = $2.45
(4) $2.45 ÷ 12 = ?
(5) $2.45 ÷ 12 = $0.204166...
(6) $0.204166 ~ $0.20
Step 1: The cost of the dozen eggs that Shinika bought can be figured by subtracting the cost of the milk from
the total bill. Remember that Shinika bought 3 gallons of milk, so multiply $2.60 by 3 and subtract that product
from 10.25.
Step 2: $2.60 times 3 equals $7.80.
Step 3: Subtract $7.80 from $10.25 to get the cost of the dozen eggs.
Step 4: The cost of the dozen eggs is $2.45. There are 12 eggs in one dozen, so divide $2.45 by 12 to determine
the cost of one egg.
Step 5: One egg costs $0.204166... Round this number to the nearest cent.
Answer: One egg costs approximately $0.20.
Example 2: Michael, Alex, Joe, and Jesse were all comparing their CD collections. Michael had 1/5 less than
Jesse. Before he bought 10 new CDs, Joe had 5 times as many as Jesse. Alex had twice as many as Joe. Joe has
110 CDs. How many CDs did Michael have?
(1) 110 - 10 = 100
(2) 100 ÷ 5 = 20
(3) 20 ÷ 5 = 4
Step 1: Determine the number of CDs that Joe had before he bought the 10 new ones.
Step 2: Before Joe bought the 10 new CDs, he had 5 times as many as Jesse, so divide 100 by 5 to determine the
number of CDs that Jesse had.
Step 3: We know that Michael has 1/5 as many CDs and Jesse, so divide 20 by 5 to determine the number of
CDs that Michael has.
Answer: 4 CDs
Irrelevant information: Alex had twice as many as Joe.
Radicals: Simplifying
A radical sign looks like a check mark with a line across the top. The radical sign is used to communicate
square roots. The numbers -3 and 3 are the square roots of 9 because both -3 and 3 squared equal 9. The
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definition of a square root is: if x squared = y, then x is a square root of y. Simplifying radicals assesses the
ability to perform operations with radicals.
In order to simplify radicals, it is important to understand the product property of square roots. We
already know the square root of 81 is 9. We can also determine this by the following method:
Simplify the square roots below:
When simplifying with variables, we can do the following:
Also, the same method can be used to simplify radicals taking the 3rd, 4th, 5th,...etc. roots.
Functions/Relations - B
A relation can be expressed as a set of ordered pairs such as
relation.
A function is a special type of
There are a few ways to determine if a relation is a function. One way is to look at all of the ordered
pairs of a relation. If no two of these ordered pairs have the same x - term (abscissa), then the relation is
a function. Another method involves graphing the ordered pairs of a relation on a coordinate graph. If
no vertical line crosses this graph at more than one point, then the relation is a function.
A linear function is a function whose graph is a line or subset of a line which is not vertical. A special
type of linear function is a constant function, whose graph is a horizontal line or subset of a horizontal
line.
Using these guidelines, find the value for q that will make the following relation not a function:
The answer is 4 because if g = 4, then the second ordered pair would be (16, 7). The two ordered pairs
would have the same x - term, and for that reason the relation would not be a function.
Inverse Variation
Variation equations are formulas that show how one quantity changes in relation to one or more other
quantities. There are four types of variation: direct, inverse (or indirect), joint, and combined. This skill
focuses on inverse variation.
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Inverse (or indirect) variation formulas show that when one quantity increases, the other quantity
decreases, and vice versa. For example, when the price of an item increases, the demand decreases.
Indirect variation formulas are of the form y = k/x, where k is the constant of variation and needs to be
determined.
Example 1: If p and q vary inversely and p = 10 when q = 1.6, find p when q = 8.
Step 1: Set up the appropriate inverse variation formula. Since p varies inversely with q, that formula is
p = k/q. Using the first set of values in the problem (because both p and q are known), substitute p and q
into the formula so that the constant k can be determined.
Step 2: Solve for k by multiplying each side of the equation by 1.6 (k = 16).
Step 3: Now that the constant k has been determined, substitute the values into the formula that are
known to determine the value of p when q = 8 and solve for p (p = 2).
Answer: p = 2 when q = 8
NOTE: A simpler way to solve inverse variation problems is to remember that the product of x and y is
always the same. The values given for x and y may change, but the product of x and y remains the same.
The following table is an example of a function in which x and y have an inverse relationship.
As stated above, the values given for x and y change, but their products remain the same. For this table,
the constant of variation, the product of x and y, is 8. Since y can always be found by dividing x into 8, y
varies inversely with x.
Once the student is comfortable with the concept of inverse variation, he or she will be ready to solve
problems in the context of real world situations.
Example 2: The amount of time it takes a seamstress to sew a wedding dress varies inversely with the
number of years of sewing experience she has. If Veronica could sew a wedding dress in 8 days when
she had 3 years of sewing experience, how many days will it take her to sew a wedding dress now that
she has 5.5 years of sewing experience? Round your answer to the nearest full day.
Step 1: Determine the constant of variation by multiplying the two terms that vary inversely (8 days and
3 years).
Step 2: Divide the constant of variation by Veronica's experience now (5.5) to determine the number of
days that it will take Veronica to sew the dress (4.3636...).
Step 3: Round 4.36 to the nearest whole day (4).
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Answer: It will take Veronica 4 days to make a wedding dress now that she has 5.5 years of sewing
experience.
As a reinforcement activity, have the student think of situations where two things vary inversely, such as
the number of years in college versus the average college student's savings account. Make up some
numbers for the various situations and solve for one variable when the other is known.
Functions: Notation
A relation is a set of ordered pairs such as
A function is a relation in which no two ordered
pairs of a set have the same x-value. Functions are often abbreviated as f(x) = y. This is read, "the value
of f at x is y."
Given y = 3x - 2, you can find the value of y when x = -3 by substituting -3 for x.
The value of y when x is -3 is -11.
The domain of a function is the set of all x-values of the ordered pairs in the function. The range of a
function is the set of all y-values. To find the range of a function when given its domain, follow the
substitution method above.
Example 1: Find the range of f(x) = 3x - 2 given that the domain is
Solution: Substitute each number in the domain into the function f(x) = 3x - 2 and simplify to determine
the numbers in the range.
f(0) = 3(0) - 2
f(0) = 0 - 2
f(0) = -2
f(1) = 3(1) - 2
f(1) = 3 - 2
f(1) = 1
f(2) = 3(2) - 2
f(2) = 6 - 2
f(2) = 4
From the domain, it can be determined that the range is
Example 2: Find f(3) + h(4), when given f(x) = 3x - 1 and h(x) = 2x + 3.
1. Substitute 3 in f(x) = 3x - 1 to get 8
2. Substitute 4 in h(x) = 2x + 3 to get 11
3. Now add f(3) + h(4) = 8 + 11 = 19
Equations: Systems
A system of equations contains at least two equations that may be linear, non-linear, or a combination of the
two types. A graphical interpretation of the solution of a system of equations is that point (or points) where the
graphs of the equations intersect. One method of finding the solution(s) of a system of equations involves
adding the two equations together.
Example 1:
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Step 1: Write the equations in a vertical format, aligning the x-terms, y-terms, equal signs, and constant
terms.
Step 2: The objective is to add the corresponding parts of the two equations together and eliminate either
the x- or y-term. Multiplying each term in the top equation by 4 will create the necessary conditions for
eliminating the y-term.
Step 3: Add like terms in the two equations, and notice the resulting y-term will have a coefficient of
zero and be eliminated.
Step 4: Solve the resulting equation for x. In this case, that means divide both sides of the equation by
14. This results with x = 5.
Step 5: To solve for y, substitute the value of x (5) in either of the original equations. The second
equation was chosen for this example.
Step 6: Subtract 10 from both sides of the equation.
Step 7: Solve the resulting equation for y. In this case, that means divide both sides of the equation by 4. This results with y = -2.
As with all other equations, substitute the values of x and y into the original equations to ensure they are
correct solutions.
Both values check out. Therefore, the solution to the system of equations are x = 5 and y = -2. This can
be interpreted as the ordered pair (5, -2) of the point of intersection of the graphs of these two equations.
Equations of a Line
Every line on any coordinate graph has a corresponding equation which describes every point on the line.
Every linear equation (equation of the line) contains a slope. The slope of a line is the same between any two
points on the line.
Before you can find the equation of a line, you must first be able to find the slope of a line when given
two coordinate points on the line. These two points are named:
The formula for the slope of
a line follows.
Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).
Step 1: Substitute the given coordinate points into the formula.
Step 2: Simplify the fraction.
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Answer: The slope of the line is 1.
The Point-Slope form for the equation of a line:
Example 2: Use the following points to find the equation of the line.
Point T (7, -3)
Point U (-4, 6)
Step 1: Solve for the slope of the line between Point T and Point U.
Step 2: Use one of the coordinate points and the slope and substitute them into the Point-Slope form for
the equation of a line.
Step 3: Simplify both sides of the equation.
Step 4: Subtract 3 from both sides of the equation.
The equation of the line that passes through (7, -3) and (-4, 6) is y = -9/11x + 30/11.
Number Relation Problems
Number relation problems involve sentences that must be translated into equations.
To translate sentences, review the following phrase/number equivalences:
If x is equal to 4, then x = 4
If the sum of 2x and 2y is 23, then 2x + 2y = 23
If 3 is less than twice y, then 2y > 3
If 5x decreased by y is more than 2y, then 5x - y > 2y
If the result of 7 more than 3 times x is y, then 3x + 7 = y
Example: One number is 2 less than another number. If twice the larger number is decreased by 3 times
the smaller number, the result is 20.
Step 1: Translate the first sentence in the problem into the first equation.
Let x and y represent the unknown numbers, with y representing the larger number and x representing
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the smaller number.
Step 2: Translate the second sentence in the problem into the second equation.
Use x and y as in the first equation, substituting them into their respective locations in the equation.
Step 3: Substitute x + 2 for y in the second equation.
Step 4: Distribute the 2 across (x + 2).
Step 5: Combine the similar terms.
Step 6: Subtract 4 from both sides.
Step 7: Divide each side by -1 to find the value of x.
Step 8: Using x = -16, substitute for x in the second equation.
Step 9: Multiply -3 and -16.
Step 10: Subtract 48 from both sides.
Step 11: Divide each side by 2.
Answer: x = -16 and y = -14
As always, check the solutions by substituting both equations with x = -16 and y = -14.
Factoring
Consider the following equation:
3 x 4 = 12
The numbers 3 and 4 are said to be factors of the number 12. This concept of factoring is not reserved for
numbers, but may be extended to polynomials as well.
Factoring is the breaking up of quantities into products of their component factors. One way to think of
factoring is as the opposite or inverse of multiplying.
A polynomial is a term or sum of terms. Each term is either a number or a product of a number and one or more
variables.
A monomial is a polynomial with one term.
A binomial is a polynomial with two terms.
A trinomial is a polynomial with three terms.
Consider the following polynomial:
A typical question on factoring will include a polynomial like the one above. Notice that 4y is a
common factor of each term of the polynomial.
Step 1: Factor out the 4y by dividing each term of the trinomial by 4y.
Step 2: The trinomial in parentheses can be factored further. Since the coefficient of the "y squared"
term is equal to 1, focus on the last term, in this case, -5. If factors of -5 can be found that ADD up to the
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coefficient of the middle term (in this case, 4) the trinomial can be factored. Two factors of -5 are
5 and -1, and when ADDED together the result is equal to the coefficient of the middle term, 4. Notice
how these numbers are put together to construct the fully factored trinomial:
(4y)(y + 5)(y - 1)
To check the result, use the rules for multiplying polynomials and you should have the original
polynomial when finished.
Sometimes factoring must be done by grouping. The polynomial given below may appear impossible to
factor at first, but if you examine the steps you will see a method to use with polynomials of this type.
Step 1: (5x) is a common term to both 15x and 20xy. Factor it out of only those two terms.
Step 2: (6n) is a common term of both 18nx and 24ny. Factor it out of only those two terms.
Step 3: Notice that the quantity (3x + 4y) is a common factor of 5x and 6n. The expression is rewritten
to indicate this, and the polynomial is completely factored.
A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type
are factored easily once the rule is remembered.
Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6
respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6).
Notice that if these terms are multiplied together, the original polynomial is formed. Polynomials that
are in the "difference of squares" form may always be factored as the sum of the square roots times the
difference of the square roots.
Quadratic Formula
A quadratic equation is a polynomial equation in which the highest power of the unknown variable is two.
An example of a quadratic equation is below.
The format of a quadratic equation is
Quadratic equations can be solved by factoring,
graphing, or by using the quadratic formula. The quadratic formula is as follows:
It can be found in any algebra textbook. This formula should be memorized.
To apply the formula to a quadratic equation, use the quadratic equation format given above as a
guideline.
Example 1: Solve the quadratic equation.
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Step 1: Determine the values of a, b, and c and substitute them into the quadratic formula. a = 1, b = 6,
and c = -91
Step 2: Determine the value under the radical symbol. 6 squared is 36 and -91 times -4 equals 364.
36 + 364 = 400
Step 3: The square root of 400 is 20 (20 x 20 = 400).
Step 4: Split the remaining problem into two problems: (-6 + 20) ÷ 2 and (-6 - 20) ÷ 2 and solve the two
problems.
The answers are x = 7 and x = -13.
Example 2: Solve the quadratic equation.
Step 1: Write the equation.
Step 2:
This will put the equation in standard form.
Step 3: Determine the values of a, b, and c.
Step 4: Substitute the values of a, b, and c into the quadratic formula.
Step 5: Determine the value under the radical sign. The square root of 0 is 0.
Step 6: Solve for x.
Answer: x = -2
The discriminant is the portion of the quadratic equation under the radical sign
properties below will give you vital information about quadratic equations.
The discriminant
1. If the discriminant is a perfect square, then the quadratic equation can be factored.
2. If the discriminant is greater than 0, then the equation has two real solutions.
3. If the discriminant is less than 0, then the equation has no real solutions.
4. If the discriminant is equal to 0, then the equation has one real solution.
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Example 3: How many solutions does the following quadratic equation have?
Step 1: Determine the values of a, b, and c.
Step 2: Substitute the values for a, b, and c into
Step 3: Simplify the discriminant.
Since the discriminant is greater than zero, there are two real solutions.
If the discriminant is a perfect square, then the solutions are rational.
If the discriminant is not a perfect square, then the solutions are irrational.
Polynomials: Addition
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
To add polynomials, combine similar terms.
Example 1:
Step 1: Set the two polynomials up as one long polynomial.
Step 2: Combine the like terms.
Step 3: Add the results of combining the like terms to determine the answer.
The sum is
Example 2:
Step 1: The area of the large rectangle would be the area of the shaded region added to the area of the
small rectangle. Since we know both areas, we simply add them together.
Step 2: Collect like terms so they can be added.
Step 3: Add together any like terms that were collected to determine the final answer.
The area of the large rectangle is
Sometimes it is necessary to use the distributive property before we can combine like terms.
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Example 3:
Step 1: Multiply each term of the first polynomial by 3. Then multiply each term of the second
polynomial by -4.
Step 2: Group like terms together.
Step 3: Combine like terms.
Answer:
Example 4: Solve for a, b, and c.
Step 1: Group the similar terms on the left side of the equation together.
Step 2: Now, group like terms from both sides of the equal sign together.
Step 3: Solve for a, b, and c.
Answer: a = -4, b = -2, and c = 1
Exponential Notation - E
Exponents communicate the number of times a base number is used as a factor. The base number 5 to the 3rd
power (with an exponent of 3) translates to 5 x 5 x 5. (5 to the 3rd power is not 5 x 3.) The result of 5 to the 3rd
power is 125. To perform operations with exponents, all exponential properties must be understood.
To simplify an expression or to find the missing term in a simplified expression, apply the following
exponential properties which are listed below.
Product of Powers:
When multiplying two (or more) numbers with the same base that have exponents, the base remains the
same and the exponents are added.
Power to a Power:
When taking a number with an exponent to another power, the base remains the same and the exponents
are multiplied.
The Power of Zero:
Any number taken to the power of zero (except zero) equals 1.
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Negative Exponents:
There is a rule for evaluating negative exponents.
Example 1: Simplify.
Step 1: Separate the expression into products of whole numbers times products of variables with like
bases. A variable or number that is written without an exponent automatically has an exponent of 1.
Step 2: First, multiply the whole numbers to get 6. Use the "Product of Powers" rule to evaluate the
terms in the second set of parentheses and the terms in the third set of parentheses.
Step 3: Add the exponents to complete the expression.
Answer:
Example 2: Simplify.
Step 1: Separate the expression into products of each of its terms. The exponent (2) is given to each term
inside the parentheses.
Step 2: -2 to the second power is (-2)(-2). Use the "Power to a Power" rule to evaluate the rest of the
expression.
Step 3: Multiply the powers to determine the exponents on the variables. The final expression can be
written with the parentheses still in it, or without the parentheses. It is more acceptable to remove the
parentheses.
Polynomials: Multiplication
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a
polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.
Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are
monomials that have the same variable or variables for which the variable or variables have the same
exponent.
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To multiply monomials and polynomials, the exponential properties must be followed. These properties
apply to all real numbers with positive exponents.
Exponential property #1 can be used to obtain results for problems such as:
Exponential properties #2 and #3 can be used to obtain results such as:
To multiply a monomial and a polynomial, multiply the monomial by each term of the polynomial. In
other words, distribute the monomial.
To multiply two polynomials, each term of the first polynomial must be multiplied by each term of the
second polynomial. Distribute each term of the first polynomial across the second polynomial.
Example 1: Which choice below represents the following trinomial:
We know that when multiplying two polynomials, each term of the first polynomial must be multiplied
by each term of the second polynomial. Let's look at our solutions.
Answer: B
Example 2: Given these measurements, find the area of the rectangle.
Step 1: Area = Length x Width. Determine the length and width of the rectangle.
Step 2: Multiply the length and width to determine the area of the rectangle.
Step 3: Use FOIL to multiply the two binomials. Remember, multiply the first terms in each binomial,
then multiply the outer terms, next multiply the inner terms, and finally multiply the last terms.
Step 4: The final step in solving this problem is to add the like terms. 3n + 7n = 10n
Answer:
Polynomials: Division
A monomial is the product of a number and an unknown variable or unknown variables. 6xy is a
monomial. The sum of two or more monomials is called a polynomial. Here is an example of a
polynomial:
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A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a
polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.
Before dividing polynomials, recall the following properties associated with exponents:
Example 1: Divide.
Step 1: Divide the whole numbers: 12 ÷ -3 = -4.
Step 2: Use the properties above to divide the variables. Begin with the x-variables. x-cubed divided by
x equals x-squared.
Step 3: Now divide the y-variables. y divided by y equals y to the power of zero. Any number taken to
the power of zero equals 1.
Step 4: Finally, multiply the quotients back together.
The answer is
Dividing a Polynomial by a Monomial:
To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Then,
combine the similar terms.
Example 2: Divide.
Step 1: Divide 3m by 3, to get m.
Step 2: Divide -9n by 3, to get -3n.
Step 3: Combine the terms.
Answer: m - 3n
Dividing a Polynomial by a Polynomial:
Dividing one polynomial by another is very similar to long division.
Example 3:
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Step 1: Write the problem as a long division problem. The binomial belongs on the outside of the
division symbol because it is the term we are dividing by.
Step 2: Now, we can begin dividing.
So, 2x belongs above the 8x.
Step 3: The next step is to multiply 2x by (3x + 1).
Subtract that product from
Now, bring the + 8 straight down beside the
6x.
Step 4: (3x)(2) = 6x, so we place the 2 above the 8 in the answer.
Step 5: Multiply 2 by (3x + 1) to get 6x + 2. Subtract (6x + 2) from (6x + 8). There is a remainder of 6,
so we write the remainder as a fraction with the binomial as the denominator.
Answer:
Expressions: Evaluating & Simplifying
Expressions look like equations except expressions do not have equal (=) signs. Expressions are "evaluated" or
"simplified," not "solved."
To evaluate and simplify expressions with brackets and parentheses, use the rules regarding Order of
Operations, Integer Properties, and Evaluating.
Here is the order of operations:
(1) Parentheses, Brackets, and Braces
(2) Exponents or Roots
(3) Multiply or Divide in order from left to right
(4) Add or Subtract in order from left to right
The order of operations is the same whether you are working with whole numbers, fractions, or
decimals. Here are a few helpful hints for using the order of operations. First, remember to complete all
operations of one type before moving on to the next type (for example, complete all multiplication and
division before moving on to addition or subtraction). Second, remember that when working the
multiplication or division move from the left to the right (for example, 2 x 6 ÷ 3. In this case, you would
multiply first because the multiplication is the first operation when reading from the left to the right).
Finally, addition and subtraction work the same way as multiplication and division - from the left to the
right (for example, 10 - 6 + 2. In this case, you would subtract 6 from 10 first, then add the 2).
Evaluating an expression can occur in two ways. If the expression contains only operations and
numbers, simply perform the indicated operations, observing the correct order. If the expression contains
variables as well, number values will have to be substituted for each variable, and then simplified using
the correct order of operations.
Example 1: Evaluate 3x + 2 if x = -2
(1) 3(-2) + 2
(2) -6 + 2
(3) -4
Step 1: Substitute -2 into the expression in place of the x's.
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Step 2: Use the order of operations and multiply 3 and -2 first.
Step 3: Complete the addition.
Some expressions may have two variables. If values are given for both variables, substitute each value
into their respective variable within the given expression.
Example 2: Evaluate 3(x - 3y) if x = -2 and y = 2
(1) 3(-2 - 3(2))
(2) 3(-2 -6)
(3) 3(-8)
(4) -24
Step 1: Substitute the values of x and y into the expression.
Step 2: Perform the multiplication that is within parentheses first.
Step 3: Subtract the numbers inside the parentheses.
Step 4: Multiply to complete the problem.
Example 3:
Step 1: Substitute the values given for x and y wherever x and y are in the expression.
Step 2: Simplify within the parentheses.
Step 3: Multiply within the brackets.
Step 4: Add within the brackets.
Answer: -4
Radicals: Equations
Radical equations are equations in which there are numbers and or letters inside the radical sign. The numbers
and/or letters are called the radicand. The following is an example of how to solve a radical equation.
Example 1: Find the solution set of the radical equation.
Step 1: Combine the whole numbers by subtracting 4 from both sides.
Step 2: Square both sides.
Step 3: Simplify each side of the equation.
Step 4: Add 3 to both sides of the equation.
Step 5: Divide both sides of the equation by 4.
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The answer is: x = 3.
Substitute the result into the original equation to verify that it is correct.
Example 2: The square root of the quantity of 3 times a number increased by 4, equals 11. Find the
number.
Step 1: Write the equation.
Step 2: Square both sides of the equation to get rid of the square root.
Step 3: Simplify both sides of the equation.
Step 4: Subtract 4 from both sides of the equation.
Step 5: Divide both sides of the equation by 3.
The answer is: x = 39.
Radicals: Addition/Subtraction - B
A radical sign looks like a check mark with a line across the top. The radical sign is used to communicate
square roots. Students must be able to simplify and combine radicals with similar terms.
To add or subtract radicals, the numbers and variables under the radical sign (known as the radicand)
must be identical. The square root of 3 and the square root of 27 can not be added together in their
present form because their radicands (in this case, 3 and 27) are not identical. However, the square root
of 27 can be simplified to 3 multiplied by the square root of 3. The two terms can now be added together
because both radicands are equal to 3. Taking the square root of 3 and adding it to 3 multiplied by the
square root of 3 results in 4 multiplied by the square root of 3 in the same way that x + 3x = 4x.
Example 1: Subtract.
Step 1: Simplify the terms under the radicals by finding the largest possible perfect square number that
will divide into 48 (16) and the largest possible perfect square number that will divide into 27 (9). Then
simplify the variables that are under the radicals by separating out any square terms.
Step 2: Replace 16 with 4 squared and 9 with 3 squared.
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Step 3: Simplify the first radical by taking a 4 and an |m| out and multiplying them by the 2 already in
front of that radical. Simplify the second radical by taking a 3 and an |m| out and multiplying them on
the outside of the radical. The m is in an absolute value symbol because we cannot have a negative
answer when we take the square root and taking the absolute value removes any possibility of a negative
answer.
Step 4: Rewrite the radicals after completing all multiplications.
Step 5: Now, subtract 3|m| from 8|m| and leave the terms under the radicals alone.
Step 6: Write the difference of 8|m| and 3|m| in front of the radical.
*Remember, when simplifying that m + |m| does not equal 2m because m and |m| are not like terms.
Example 2: Find the approximate square root to the nearest tenth for each term. Then combine the
terms.
(1) 2.2 + 6.8 +9.4 - 7.3
(2) 2.2 + 6.8 +9.4 - 7.3 = 11.1
Step 1: Take the square root of each term using a square root chart or your calculator. Round each term
to the nearest tenth.
Step 2; Combine like terms.
Example 3: Add.
Step 1: Simplify the terms under the radicand by finding the largest cubed number that will divide into
192 (which is 64). Do the same for 375 ( it is 125).
Step 2: Simplify the first radical by taking out the cubed root of 64 (4) and multiplying it by 3. Simplify
the second radical by taking out the cubed root of 125 (5) and multiplying it by 2.
Step 3: Rewrite the radicals after completing all multiplications.
Step 4: Add 12 and 10 together since the radicands are the same. The radicands do not get changed.
Answer:
Radicals: Multiplication/Division
A radical sign looks like a check mark with a line across the top. The radical sign is used to communicate
square roots. The skill Multiplying/Dividing Radicals evaluates the ability to find the product or quotient of 2
or more radicals.
To multiply radicals, multiply the numbers in front of the radical sign and the numbers which make up
the radicand (the number or numbers under the radical sign).
Example 1:
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Step 1: Identify the expressions to be multiplied.
Step 2: Multiply the numbers in front of the radical sign.
Step 3: Multiply the radicands, leaving the product as a radicand.
The answer is
When dividing radicals, the problem can be set up as a fraction. If the denominator of this fraction
contains a radical, the denominator must be rationalized. To do this, multiply the numerator and
denominator by the radical found in the denominator. As a result, the numerator may now contain a
radical, but the denominator will not. Simplify the numerator, and then simplify the entire fraction if
possible, and the process is complete.
Example 2:
Step 1: Multiply the numerator and denominator by the radical to leave a whole number in the
denominator.
Step 2: Simplify by dividing the 7 into 63.
Step 3: Look for perfect squares factors.
Step 4: Simplify. A radical is simplified when it contains no perfect squares.
Example 3: Multiply the expressions.
Step 1: Multiply the first two terms, the outer two terms, the inner two terms, and the last two terms.
Step 2: Multiply the numbers outside the radicals, then multiply the numbers inside the radicals of each
term. Simplify, if possible.
When rationalizing the denominator of a fraction, multiply the numerator and denominator by the
conjugate of the denominator.
Example 4: Rationalize.
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Step 1: Multiply the numerator and denominator by the conjugate of the denominator.
Step 2: Begin to simplify the numerator by multiplying radical 5 and 2 by 5. Begin to simplify the
denominator by multiplying the binomials.
Step 3: Complete all multiplications, then add all like terms.
Step 4: Complete the simplification of the denominator by subtracting 4 from 5.
Step 5: Divide both terms in the numerator by the number in the denominator.
Answer:
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