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ALGEBRA REVIEW
Translating Verbal Expressions into Mathematical Expressions
Verbal Expressions Examples
Math Translation
Addition
added to
more than
the sum of
increased by
the total of
6 added to y
8 more than x
the sum of x and z
t increased by 9
the total of 5 and y
6+y
8+x
x+z
t+9
5+y
Subtraction
minus
less than
subtracted from
decreased by
the difference
between
x minus 2
7 less than t
5 subtracted from 8
m decreased by 3
the difference between y and 4
x-2
t-7
8–5
m-3
y-4
10 times 2
one half of 6
the product of 4 and 3
10 X 2
(1/2) X 6
4X3
multiplied by
y multiplied by 11
11y
divided by
the quotient of
x divided by 12
the quotient of y and z
x/12
y/z
the ratio of
the ratio of t to 9
t/9
Power
the square of
the cube of
squared
the square of x
the cube of z
y squared
x2
z3
y2
Equivalency
equals
is
is the same as
1+2 equals 3
2 is half of 4
½ is the same as 2/4
1+2 = 3
2 = (½)X4
yields
represents
3+1 yields 4
y represents x+1
3+1 = 4
y=x+1
greater than
less than
greater than or equal
to
at least
no less than
less than or equal to
at most
no more than
-3 is greater than -5
-3 > -5
-5 is less then -3
-5 < -3
x is greater than or equal to 5 x ≥ 5
Multiplication times
of
the product of
Division
1
2

2
4
Comparison
x is at least 80
x is no less than 70
x is less then or equal to -6
y is at most 23
y is no more than 21
x ≥ 80
x ≥ 70
x ≤ -6
y ≤ 23
y ≤ 21
WORD PROBLEMS
Setting up word problems:
1) Find out what you are being asked to find. Set a variable to this unknown quantity. Make sure you know the
units of this unknown (miles?, hours? ounces?)
2) If there is another unknown quantity, use the given information to put that unknown quantity in terms of the
variable you have chosen.
(For example, if total distance traveled is 700 miles, then part of the trip is x miles and the other part of the trip
is 700 – x miles.)
3) Set up a table with a row for each unknown and columns made up of the terms of one of equations above (r*t =
d, Pr = I, etc..)
4) Use the given information to combine the equations of each row of the table into one equation with one variable
to solve for.
5) Check your equation by plugging in your value for x and seeing if your equation is true.
6) Once one variable is solved for, you can find the other unknown. (For example, is x = 100 miles, then the other
part of the trip is 700 – 100 = 600 miles.)
7) Check your equation by plugging in your value for x and seeing if your equation is true.
Distance Problems
Distance = rate * time
If distance is in miles, and rate is in mph, then time must be in hours.
If time is in minutes, then multiply time by 1 hr/60min to convert to hours.
PERCENT = part of a hundred
5% =
5
 .05
100
When converting from Percent to a Decimal, drop the % sign and move the decimal place to the
LEFT TWO PLACES.
635% = 6.35
When converting from a Decimal to a Percent, move the decimal place to the RIGHT TWO PLACES and then attach a % sign
to the end.
.076 = 7.6%
When converting a fraction to a percent, first convert the fraction to a decimal (see previous page).
Percent Problems:
Amount = Percent * Base
Amt
4
is 40% of 10
t
Example 1:
What is 30% of 60?
Amt = Percent X Base
Percent
Amt = 30% X 60
(in dec form)
Convert 30% to decimal.
30% = .30
Amt = .30 X 60 = 20
Example 2:
200% of what is 400?
200% is the Percent
Converting Percent to a Decimal gives 200% = 2
400 is the Amount
“What” is the Base.
Base = Amt / Percent = 400 / 2 = 200
Base
Example 3:
3 is what percent of 45?
Amt = 3
Base = 45
Percent = Amt/ Base
Percent = 3/45 = 0.06666…..
Converting to Percent using % symbol gives
6.66666…% which rounds to 6.67%
WORK
Rate of Work * Time Worked = Part of Tasked Completed
If someone can do a job in 60min, their rate of work is 1 joh/60min.
If someone else can do the same job in 40minutes, their rate of work is 1 job/40min.
The TIME to get the same job done TOGETHER can be found by
Adding their parts together to make 1 whole job.
Similar Triangles
Triangles are similar if at least 2
corresponding angles are the same
in each triangle.
Pythagorean Theorem
The longest side of a right triangle is called the "hypotenuse", so the formal definition is:
In a right angled triangle the square of the hypotenuse is equal to
the sum of the squares of the other two sides.
So, the square of a (a²) plus the square of b (b²) is equal to the square of c (c²):
a2 + b2 = c2
FRACTIONS
A fraction is just a division problem.
Numerator
= Numerator ÷ Denominator
Denominato r
A fraction is in LOWEST TERMS when the numerator and denominator have no common factors.
6
3
is not in lowest ter ms,
is in lowest ter ms
10
5
3
Mixed Number - A number, such as 6 , consisting of an integer and a fraction. A mixed number is just the sum of a
5
whole number and a fraction.
Improper Fraction - A fraction in which the numerator is larger than the denominator. Converting from mixed number to
improper fraction:
3 6  5  3 33
6 

5
5
5
Mixed numbers must be converted to improper fractions or decimals before doing ANY
MULTIPLICATION OR DIVISION OPERATIONS on them.
A fraction can be converted into a decimal by dividing: Numerator ÷ Denominator
denominato r numerator
When ADDING or SUBTRACTING FRACTIONS, they must have the same denominator, then
you just add the numerators and leave the denominator the same.
1 3
 
6 8
The denominators, 6 and 8, are not the same, so we must find the LEAST COMMON DENOMINATOR
to convert these fractions into equivalent ones with the same denominator. The LCM of 6 and 8 is
the SMALLEST NUMBER THAT BOTH 6 and 8 can go into. Choose the larger denominator
(which is 8 in this case) and start taking multiples until 6 can go into it.
Does 6 go into 8? NO
Does 6 go into 8x2? NO
Does 6 go into 8x3? Yes, 6 goes into 24. Therefore 24 is the LCM.
Multiply the numerator and denominator of each fraction by whatever it takes to get the LCM as
the new denominator.
Then once you have the same denominators in each fraction, just add the numerators and leave the
denominator the same.
1  4 33 4
9 13


   
6  4  8  3  24 24 24
MULTIPLYING FRACTIONS
Multiplying Fractions uses a different rule. When multiplying fractions, you multiplying the numerators
AND the denominators.
3
 1  3  1  3

   
 6  8  6  8 48
This fraction can simplified by canceling out common factors in the numerators and denominators (tops
and bottoms)before multiplying across. 6 can be rewritten as 2x3, and the 3’s can be cancelled out.
1 3
1
 1  3  1  3


   
 6  8  6  8 2  3  8 16
DIVIDING FRACTIONS
The rule for dividing fractions is simple. Just take the RECIPROCAL(flip the top and bottom) of the
DIVISOR (the fraction to the right of the ÷ symbol) multiply
 1   3   1   8  1 8 1 2  4 4


        
 6   8   6   3  6  3 2  3 3 9
GEOMETRY
x- intercept - point where graph crosses x-axis. y = 0 at this
point.
y-intercept – point where graph crosses y-axis. x = 0 at this
point.
slope = rise/run of a line, aka change in y/change in x. If linear
equation is in the form
The slope-intercept form of a linear equation is y = mx + b,
where m is the slope and ( 0 ,b ) is the y-intercept.
If two points of a line are known, (x1,y1) and (x2,y2), the slope
can be found with this equation, the slope,
y 2  y1
m=
x2  x1
The point slope formula for a line with point
(x1, y1) and slope m is y – y1 = m(x – x1)
y= - ½ x +1
ORDERED PAIR:
The first number in the ordered
pair is the x-coordinate and the
second number in the ordered
pair is the y-coordinate.
Horizontal Lines = lines parallel to the x-axis with an equation in the form y = constant.
Vertical Lines = lines parallel to the y-axis with an equation in the form x= constant.
Parallel lines – have the same slope.
Perpendicular lines- form right angles. Their slopes are negative reciprocals of each other. Example:
y=2x+1 and y= -½ x - 2 are perpendicular.
To graph a LINEAR INEQUALITY,
First rewrite the inequality to solve for y.
If the resulting inequality is y > ….,
Then make a dashed line and shade the area ABOVE the line.
If the resulting inequality is y < …..,
Then make a dashed line and shade the area BELOW the line.
If the resulting inequality is y≥ …..,
Then make a SOLID line and shade the area ABOVE the line.
If the resulting inequality is y ≤ …….,
Then make a SOLID line and shade the area BELOW the line.
If when isolating y, you must divide both sides of inequality by a negative number,
Then the inequality sign must be SWITCHED.
Example: -3y + 6x < 12
-3y < =6x + 12
divide through by -3 and switch inequality sign
y > 2x - 4 Now graph a dashed line with y-intercept (0,4) and slope 2 and shade above the line.
Perimeter – distance around the edges of an object
Perimeter of a square = 4s, where s = length of one side
Perimeter of a rectangle = 2W + 2L, where W = width and L = length
Perimeter of a triangle = side 1 + side 2 + side 3
Area – amount of surface covered by an object.
Area of a square = s2
Area of a rectangle = L*W
Area of a triangle = ½ bh, where b = length of base and h = height
INTEGERS
The set of natural numbers is {1,2,3,4,5,6,7,….}. These are basically the “counting numbers.”
The set of whole numbers is the set of natural numbers and the number, 0. {0, 1, 2, 3, 4, 5, ….}
The set of integers is the set of whole numbers and their opposites. {…. -4,-3,-2,-1,0,1,2,3,4,….}
The number 0 is an integer, but it is neither negative nor positive. For any two different places on the
number line, the integer on the right is greater (>) than the integer on the left.
The absolute value (using the | | symbol) of a number is its distance from zero on the number line. The
absolute value of a number is ALWAYS POSITIVE (or 0).
| 5| = 5,
|-5| = 5,
|3-8| = |-5| = 5, |3-3| = |0| = 0
|-4|=4
-4 is 4 units away
from 0
|4|=4
+4 is 4units away
from 0
Adding and Subtracting Integers
When adding two integers with the same sign,
just ignore the signs, then attach them on the
answer.
-3 + -5 = - (3 + 5) = -8
When adding two integers with different signs,
take the absolute values (make both numbers
positive), and subtract the smaller one from the
larger one. The sign of the integer with the
larger absolute value will be the sign of your
answer.
7 + (-8)
The absolute value of 7 is |7| = 7
The absolute value of -8 is |-8| = 8 larger
8–7=1
Remember in the original problem, the integer
whose absolute value was 8 was -8, so our
answer is negative.
7 + (-8) = -1
Subtracting a negative integer is the same as
________ a _______ integer.
3 – (-5) = 3 + 5 = 8
-2 – (-3) = -2 + 3 = 1
Subtracting a positive integer is the same as
adding a negative integer.
5 – 3 = 5 + (-3) = 2
-2 – 3 = -2 + (-3) = 1
Multiplying and Dividing Integers
(positive integer) X (positive integer) = (positive integer)
(positive integer) ÷ (positive integer) = (positive integer)
(negative integer) X (negative integer ) = (negative integer)
(negative integer) ÷ (negative integer ) = (negatrive integer)
(positive integer) X (negative integer) = (negatve integer)
(positive integer) ÷ (negative integer) = (negative integer)
(negative integer) X (positive integer ) = (negative integer)
(negative integer) ÷ (positive integer ) = (negative integer)
EXPONENTS & POLYNOMIALS
POLYNOMIALS
Exponent - A number or symbol, as 3 in (x + y)3, placed to the
right of and above another number, variable, or expression
(called the base), denoting the power to which the base is to be
raised. Also called power.
The exponent (or power) tells how many times the base is to be
multiplied by itself.
Example 1:
(x + y)3 = (x + y)(x+y)(x+y)
Example 2:
(-3)4 = (-3)(-3)(-3)(-3) = 81
Properties of Exponents
If m and n are integers, then
x m x n  x mn
If m and n are integers, then
m n
mn
(x )  x
If x is a real number and
then
x  0,
x0  1
If m, n, and p are integers, then
( xy ) n  x n y n , and
( x m y n ) p  x m p y n p
Rational Expoents
am/ n = n am
polynomial – is a term or sum of terms in which all variables
have whole number exponents. Example: 3x, or x2 + 1, or -3x2
+ 3x + 1
monomial – a number, a variable, or a product of numbers and
variables.
Example: 3, 2x, -4x2 are all monomials.
binomial – the sum of two monomials that are unlike terms.
trinomial – the sum of three monomials that are unlike terms.
like terms – terms of a variable expression that have the same
variable and the same exponent.
Example: 3x and 3x2 are unlike terms, but 3x and 2x are like
terms.
factor – (in multiplication) a number being multiplied.
If m and n are integers and x  0 , Example: What are the factors of 121? 1, 11, and 121.
then
121 = 11 X 11, 121 = 1 X 121
to factor a polynomial – to write a polynomial as a product of
m
m n
other polynomials
to factor a trinomial of the form ax2 + bx + c - to express the
n
trinomial as the product of two binomials.
Example: x2 + 5x + 6 = (x+2)(x+3)
to factor
grouping
x
0 , then – to group and factor terms in a
If n is a positive integer (-n is negative)
, and by
polynomial in such a way that a common binomial
1
1
factor is found.
x  n  n and  n  x n
Example: 2x(x+1) – 3(x+1) = (x + 1)(2x – 3)
x
x
factor completely – to write a polynomial as a product of factors
that are nonfactorable over the integers.
If n is an integer and b  0 , then
FOIL method – A method of finding the product of two
n
binomials in which the sum of the products of the First
an
 a
   n
terms, of the Outer terms, of the Inner terms, and of the Last
 b
b
terms is found.
Example:
If n is a positive integer (-n is negative)
, and b(x+2)(x+3)
 0 , then =x2+ 3x + 2x + 2*3 = x2 + 5x + 6
common
factor
– a factor that is common to two or more
n
n
numbers.
 a
 b
   
Example: What are the common factors of 12 and
 b
 a
16x2?
The factors of 12x are 1,2,3,4,6,12, and x
The factors of 16x2are 1,2,4,8,16, x, x
The common factors of 12x and 16x2 are 1x, 2x, and 4x
The Greatest Common Factor of 12x and 16x2 is 4x.
x
x
x
Perfect-Square Trinomials
(a+b) 2 = a 2 + 2ab + b 2
(x+3)2 = x2 + 6x + 9
Difference of squares
a 2 - b 2 = (a+b)(a-b)
4x2 – 25 = (2x + 5)(2x – 5)
Rules for Variable Expressions:
Only like terms can be added, and when adding like terms, do not
change the exponent of the variable.
5x2 + 3x2 = 8x2
When multiplying variable expressions, add exponents of like variables
(5xy3)(2y2)=10xy3+2 = 10xy5
When taking powers of variable expression that is a monomial (one
term), multiply exponents of EVERY term inside the parentheses.
(2x3y4)3 = 23x3*3y4*3 = 8x9y12
When taking powers of a variable expression that is a binomial, trinomial
or some other polynomial, use the rules of polynomial multiplication.
For example: (x+2)2 ≠ x2 + 22
(x+2)2 = (x+2)(x+2) = x2+4x + 4 (FOIL METHOD)
Example 2:
(x2+3x+5)2 = (x2+3x+5)(x2+3x+5)
=(x2+3x+5)x2 + (x2+3x+5)3x + (x2+3x+5)5
(DISTRIBUTIVE PROPERTY)
25 x 2 y 3 5 x

3y
15 xy 4
A General Strategy for Factoring a Polynomial
1.
Do all the terms in the polynomial have a common factor? If
so, factor out the
Greatest Common Factor. Make sure that you don’t forget it in
your final answer.
Example: 24x4 - 6x2 = 6x2(4x2 - 1). Also look to see if the
other polynomial factor and be factored more. (4x2-1)=(2x1)(2x+1), so the final answer is
24x4 - 6x2 =6x2(2x-1)(2x+1),
EXPRESSIONS
EQUATIONS
2
Examples: 3+2(1-4) , - Examples: 3x + 2 = 5,
3x+x, (3x+2)2
5x + 5y=10, x(x-5) = -6
Can be simplified
Ex1: Don’t forget order of
operations!
3+2(1-4)2 can be simplified
to
3+2(-3)2 which becomes
3+2(9) which becomes
3+18 which becomes 21.
Ex2:
-3x + x can be simplified to
-2x
Can be solved:
Ex1 :
3x  5  2  3
3x 3

3 3
x 1
A dependent system is the case
when the equations represent the
same line. Therefore they intersect
everywhere on the line.
20 4

25 5
2.
After factoring out the GCF (if there is one) count the number
of terms in the remaining polynomial.
Two terms: Is it a difference of squares? Factor by using:
a2-b2 = (a+b)(a-b)
Example: 36x2 – 49 = (6x)2 – 72 = (6x-7)(6x+7)
If the polynomial can’t be factored, it is PRIME.
One solution
Ex2 :
Three terms: Is it a perfect square trinomial?
5 y  5 x  10
5 y  5 x 10


5
5
5
y  x  2
Solution is all (x, y)
that make this linear
equation t rue.
If it is it would be in the form a2x2 + 2abx + b2 ,
which is factored as (a+b)2
or a2x2 + 2abx + b2 which is factored as (a-b)2
Example: 4x2 + 12x + 9 = (2x)2 + 2(2)(3)x + 32 =
(2x + 3)2
Is it of the form x2 + bx + c?
When graphed
Factor by finding two numbers that multiply to c and
this is a line.
add to b.
Ex3 :
Example: x2 -3x - 4 = (x+1)(x-4)
because 1*-4 = -4 and 1 + -4 = -3
x( x  5)  6
x 2  5 x  6
x 2  5 x 6  0
( x  3)( x  2)  0
x  3, x  2
Two solutions
Can be evaluated: (3x+2)2 can be
evaluated at x= -1. (3(-1)+2)2 is (3+2)2 which is (-1)2 which is 1.
Can be reduced:
Examples:
Can be evaluated to see if a
solution is true.
Is (3,4) a solution of y=-x+2?
4 = -3+2=-1 NO
Systems of Linear equations have
either one solution (independent),
no solutions (inconsistent), or
infinitely many solutions
(dependent).
An independent system is the case
when the equations represent two
intersecting lines. The solution is
the intersection point.
An inconsistent system is the case
when the equations represent two
parallel lines. They never intersect.
Can’t find the numbers? Maybe the polynomial is
PRIME.
Is it of the form ax2 + bx + c?
Try factoring by the Grouping Method (or ac
Method) or Trial and Error.
Example: 2x2 + 13x + 15 (the a*c method means
multiply 2*15 which is 30.
Find factors of 30 that add up to the middle
term’s coefficient, which in this case is 13.
3*10=30 and 3+10 = 13. Split the middle term
into two parts:
2x2 + 10x + 3x + 15 and then factor by grouping.
2x(x+5)+3(x+5) = (2x+3)(x+5)
Those methods don’t work? Maybe the polynomial
is PRIME.
Four terms: Try Factoring by Grouping. Group the 1st two
terms and the last two terms. Factor out the Greatest
Common Factor from each grouping. Then factor out the
common binomial term.
3.
Always factor completely. Double check that each of your
factors can not be factored more.
4.
Check your work by multiplying the factors together.
Does it result in the original polynomial?
2
21 1
2  
5
52 5
SOLVING POLYNOMIAL EQUATIONS (2 SOLUTIONS):
Adding and Subtracting Rational Expressions –
Put Equation in standard form, ax2 + bx + c = 0.
x
4x  1
2
3 methods:
1) SQUARE ROOT METHOD: If there is no bx term or if the
equation is in the form (x+ k)2 + c = 0, then just get the constant c on
one side, take
± the square of both sides and get x by itself.
2) FACTORING METHOD: If product ac has to factors that add
up to c, then it is factorable. Factor it and use the Zero Product Product
to find the solutions. This says that if A*B=0, then A=0 or B=0.
Example: x2 -2x = -3
Standard form: x2 – 2x + 3 = 0. Factored: (x+1)(x-3) = 0
So x+1 = 0, which gives x = -1, or another possible solution is x-3 = 0,
which gives x = 3.
3) USE QUADRATIC FORMULA: If the equation ax2+ bx+ c =0 isnot
factorable, then use quadratic formula.
1
2x  x

2
Step 1: Factor the denominators, then find the LCM. The LCM of two
polynomials is the simplest polynomial that contains the factors of each
polynomial. To find the LCM of two or more polynomials, first factor
each polynomial completely. The LCM is the product of each factor the
greater number of times it occurs in any one factorization.
x
2 x  12 x  1

1
x(2 x  1)
1st denominato r : 2 x  12 x  1
2nd denominato r : x(2 x  1)
LCM  x(2 x  1)2 x  1
Step 2: Change each rational expression so that the new denominator will
be the LCM. You will multiply the numerator and denominator of each
expression by whatever it takes to get the LCM as the new denominator.
x
x
1
(2 x  1)
x2
2x 1
 



2 x 12 x  1 x x(2 x 1) (2 x  1) x(2 x  1)(2 x  1) x(2 x  1)(2 x  1)
Quadratic Formula
if ax 2 + bx + c = 0
Step 3: Add the two new fractions by adding the numerators and keeping
the denominator (the LCM) the same.
(standard form of a quadratic equation) then
x = ( -b
x2  2x 1
x(2 x  1)(2 x  1)
(b 2 - 4ac) ) / 2a
RATIONAL EXPRESSIONS
A rational expression is a fraction in which the numerator or
denominator is a variable expression (such as a polynomial). A rational
expression is undefined if the denominator has a value of 0.
A rational expression is in SIMPLEST form when the numerator and
denominator have no common factors other than 1.
6x
is not in simplest form.
9x2
2
is in simplest form.
3x
Reducing to simplest form – factor the numerator and denominator, then
cancel out any common factors in the numerator and denominator (not
common factors that are both in the numerator or both in the
denominator, e.g. side by side).
Step 4: Now factor the resulting expression and cancel out any common
factors in the numerator and denominator.
x2  2x 1
( x  1) 2

x(2 x  1)( 2 x  1) x(2 x  1)( 2 x  1)
Simplify Complex Fractions – Complex fractions are just rational
expressions with fractions within fractions. To simplify, find the LCM of
all the denominators of every fraction in the expression, then multiply the
main numerator and denominator by that LCM. Then simplify as usual.
1
1
1
1
1( x 3 )  ( x 3 )
3
x3  x 2
x 2 ( x  1)
x 
x x 
x


 x2
1
1
1
1 x3
1
1
x

1
x 1
3
3
 3
 3
(x )  3 (x )
2
2
2
x
x
x
x
x
x
1
LCD= x3
Solving Equations with Fractions – multiply BOTH
SIDES of the equation by the LCM of all denominators
in the equation. Then solve as usual.
Multiplying Rational Expressions – factor the numerators and
denominators then cancel out common factors as above, then multiply the
numerators and multiply the denominators.
Dividing Rational Expressions – change to a multiplication problem by
changing the DIVISOR into it’s RECIPROCAL.
3
x

1
LCD  x( x  1)
x x 1
3
x( x  1)   x x( x  1)   x( x  1)
x
x 1
3( x  1)  x 2  x( x  1)
If the equation is one fraction set equal to another, this
is called a PROPORTION. Solve by CROSSMULTIPLYING, then isolating the variable.
x 12

3 18
18 x  3(12)  36
18 x 36

18 18
x2