Download Section 7 Algebraic Manipulations and Solving Part 1 Expressions

Section 7 Algebraic Manipulations and Solving
Part 1 Expressions, Equations, and Inequalities: Simplifying and Solving
Before launching into the mathematics, let’s take a moment to talk about the words
expression, equation, inequality, simplify and solve. These words commonly arise in the
language of mathematics. What do they mean, and how are they different?
•
•
An equation is a mathematical statement which says that two quantities are equal. For
example,
and
are equations.
An inequality is a statement which says that one quantity is either more or less than
another quantity, or possibly equal. We use the symbols
when writing
inequalities. For example,
means a is less than b
Means a is greater than b
Means a is greater than, or equal to, b
Means a is less than, or equal to, b
When you encounter an equation or an inequality, your objective is to determine
whether it is true. For example, given the inequality
you would say it is true
because 6 is less than 7. Given the equation
you would say that it is true IF x =
5. Equations and inequalities usually involve variables. The values of the variables for
which the equation or inequality is satisfied or true are called the solutions. So, x = 5 is
the solution of the equation
.The process of finding solutions to equations and
inequalities is called solving. We never use the word solve in any other context.
•
An expression is a mathematical statement that can be simplified, but not solved. For
example,
is an expression. It is not an equation or inequality, so no solving
is involved. Instead, we would simplify this expression and get 18. Another example is
. Although this example does have a variable, this is still not an equation or
inequality, so we would not ‘solve for x’. Instead, we would simplify this expression to
get
.
During today’s meeting, you will learn some tools for simplifying expressions and then use
these tools to solve linear equations and inequalities.
Part 2 Simplifying Expressions
In order to simplify expressions like
, it is necessary to understand the order in
which operations must be performed. We discussed this in meeting 1. Sometimes, however, it
is helpful to have some additional mathematical tools. Here, we discuss three properties of real
numbers (i.e. the numbers that you find on a number line) that are useful to know when
simplifying expressions.
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For any real numbers a, b and c, the following are true:
Commutative Property of Multiplication
Commutative Property of Addition
Associative Property of Multiplication
Associative Property of Addition
Distributive Property
Of these properties, the first four are probably familiar to you already. The first two properties
tell you that you are free to add or multiply two numbers in any order. The second two
properties tell you that you are free to add or multiply three numbers in any order. So, what
about the fifth property?
What does the Distributive Property say? Well, let’s try it out. The property says, for example,
that
appears to be true.
. Is this equation true? Both the left and right sides equal 15, so it
Example 1: Rewrite each of the following expressions using the Distributive Property. The
goal is not to simplify, but to rewrite in an equivalent form. Two of them have been done for
you as illustrations.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Example 2: For each of the following expressions, the five properties above will make
simplification easier. Use the properties to simplify the expressions.
(a)
(b)
(c)
3
(d)
Now, let’s work with expressions that involve variables. Recall that quantities that are being
added or subtracted are called terms. Terms that have identical variable parts are called like
terms. For example, in the expression
which terms are like terms?
•
and
are like terms since both have .
•
and
are like terms since both have .
• -9 and 8 are like terms since neither have any obvious variable part.
Terms which lack any obvious variable part, like -9 and 8, are called constant terms. The
numbers 6, -5, 4, 2, -9 and 8 are called coefficients.
Why are like terms important to identify? It turns out that like terms can be combined and
simplified. This is done using the Distributive Property. For example, consider the expression
:
So, the expression
this is as follows:
can be simplified down to 2v. An alternate way of thinking about
6v means v + v + v + v + v + v
4v means v + v + v + v
So, if you subtract 4v from 6v, what do you have left? Canceling v’s leaves you with
v + v or 2v remaining.
Example 3: Simplify the following expressions using the Distributive Property.
(a)
(b)
(c)
(d)
4
(e)
(f)
(g)
(h)
(i)
(j)
Mathematical expressions often involve fractions. Consequently, it’s useful to be able to
manipulate fractions. The following simplifications are useful to keep in mind:
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Example 4: Use the simplifications on the previous page to find two equivalent ways of
writing each of the following expressions:
(a)
and
(b)
and
It is also useful to be able to manipulate sums/differences of fractions. Recall that fractions are
added/subtracted as follows:
Example 5: Use the above fraction property to simplify the following expressions:
(a)
(b)
(c)
(d)
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Part 3 Solving Linear Equations and Inequalities
Recall from Part 1 that solving is the process of determining when an equation or inequality is
satisfied or true, and variable values that that make an equation or inequality true are called the
solutions.
There are MANY different kinds of equations and inequalities, but today we focus only on
linear equations and inequalities, where the left and right sides are linear functions.
Example 6: For each of the following linear equations and/or inequalities, determine which of
the listed variable values is a solution.
(a)
To see which values solve the equation, we substitute each of these values into the left
and right sides of the equation to see if the two sides are equal:
Variable Value
9
0
Solution (Yes/No)
No
Yes
-6
0
(b)
To see which values solve the inequality, we substitute each of these values into the left
side of the inequality to see if the result is less than or equal to 8:
Variable Value
t = 20
Solution (Yes/No)
3
10
8
Yes
No
Yes
We begin our discussion of the solving process by considering how linear equations and
inequalities can be manipulated. We rely on the following general principle: Adding,
subtracting, multiplying or dividing both sides of an equation by the same non-zero amount
will not affect the equality of the two sides. The result is in an equivalent equation whose
solution is the same as the original equation.
For example, consider the equation
. The solution of this equation is x = 11. Adding 3
to both the left and right sides of this equation yields
or, rather,
. Notice that x =11 also solves this equation.
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Similarly, if we multiply both the left and right sides by 2, we arrive at the equation
which also has x = 11 as a solution. Check this!
What about inequalities? It turns out that the same principal holds for inequalities, with one
exception. The example below highlights the problem:
Consider the inequality
. If we multiply both the left and right sides of this inequality by
3, we arrive at the inequality
. No problem, right? Okay, now let’s try multiplying both
the left and right sides by -3. If we do this, we arrive at the inequality
. Now we have
a problem, since -6 is NOT less than -12. This resulting inequality can be corrected, however,
if we change the inequality symbol from < to > :
.
This example illustrates that extra care needs to be taken when multiplying or dividing the left
and right sides of an inequality by negative numbers. In order for the resulting inequality to be
equivalent to the original inequality and have the same solutions, the inequality symbols need
to be reversed.
Example 7: Use the above ideas to solve each of the following equations and inequalities. The
first one has been done for you as an illustration.
(a)
Our goal is to convert this equation into an equivalent equation of the form x = ___. We
do this by strategically adding, subtracting, multiplying and dividing both sides of the
equation by amounts that simplify the resulting equation, isolate all terms involving the
variable on one side of the equation, and isolate all other terms on the other side of the
equation.
In this example, the quantity on the left hand side is divided by 5. What could we do to
both sides of this equation that would simplify the left side? Let’s try multiplying both
sides of this equation by 5:
When the left side of the resulting equation is simplified, it becomes
.
Now, let’s isolate variable terms on the left side of the equation and non-variable terms
on the right side. So, we need to move the 1 to the right side of the equation. We do this
by adding 1 to both sides of the equation:
When simplified, the resulting equation is
. Recall that our goal is to convert the
original equation into an equation of the form x = ___. We are close. We currently have
, and we want to have x, or rather 1x, on the left side of the equation. To finish,
divide both sides of the equation by 2 to get
.
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When simplified, the resulting equation is
If our work above is correct,
. This is because
.
should be the solution of the original equation
. Let’s check by substituting
into the left side of the equation to see if
it equals 7:
.
(b)
Check that
is the solution by substituting it into both sides of the original
equation to see if they are equal:
(c)
(Hint: Start by simplifying the left and right sides as much as possible
first!)
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Check that
is the solution by substituting it into both sides of the original
equation to see if they are equal:
2
2
(d)
If our work above is correct, then ANY number greater than or equal to -0.6 should be a
solution. Let’s check some values:
x
Solution (Yes/No)
-2
16
No
-1
6
No
-0.6
2
Yes
0
-4
Yes
4
-44
Yes
From the table, it appears that values of x less than -0.6 are not solutions, while values
that are -0.6 or larger are solutions.
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Section 7 Homework Assignment
1. For each of the equations given, determine whether the specified variable value is a
solution.
(a)
;
(b)
;
(c)
;
2. For each of the inequalities, the range of solution values is given. Complete the
following table and then explain how the results in the table confirm the solution you
are given.
(a)
; solution:
x
0
1
2
3
4
(b)
Solution (Yes/No)
; solution:
t
-1
0
1
2
3
Solution (Yes/No)
3. Simplify each of the following expressions.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
11
(k)
(l)
4. Solve each of the following equations and inequalities. Remember to simplify each side
of the equation first!
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k) 6 – (3w+2) = 1 – w
5. How many equivalent ways can you write each of the following expressions? List as
many equivalent ways as you can think of. (Although every one of these fractions can
be written as a decimal, just focus on equivalent expressions involving fractions.)
(a)
(b)
(c)
6. Suppose a quantity is decreasing linearly at a rate of 2.3 milliliters per minute. The
present amount is 46 milliliters. Let t represent the number of minutes elapsed since the
present time (independent variable) and let A represent the quantity remaining
(dependent variable).
(a) Determine an equation relating the variables t and A.
(b) Use your function from part (a) to determine the amount remaining after 15
minutes.
(c) Use your equation to determine how many minutes will pass before half of the
present quantity is gone. You’ll have to solve an equation.
(d) How long will it take before there are 5 milliliters remaining?
7. Suppose the population of a country is growing linearly at a rate of 634,000 people
each year. The current population of the country is 56 million. Let t represent time (in
years since the current year) and let P represent the population of this country.
(a) Determine an equation relating the variables t and P.
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(b) Use your function from part (a) to project the size of the population in the year
2020.
(c) Use your function to determine the projected number of years until the population
of this country reaches 60 million.
(d) How long will it take for the population of this country to double in size from its
current population?
8. Recall example 9 from meeting 6 which looked at monthly telephone costs.
(a) Write down the equation which related the minutes used each month and the total
monthly cost. Write down what the independent and dependent variables are, and
give their letter names. Use complete sentences.
(b) Write down an equation that you would have to solve in order to determine the
number of minutes you can use in a month if you want to spent exactly $50 for that
month. Solve this equation and interpret your solution in the context of this
problem.
(c) Write down an inequality that you would have to solve in order to determine the
range of monthly minutes you can use if you want to spend at most $70 each month.
Solve this inequality and interpret your solution in the context of this problem.
9. Use the linear functions that you created in exercises 7 and 9 from Meeting 6 to answer
the following questions?
(a) In what year is the population of the world projected to reach 7.5 billion people?
(b) In what year is the number of internet servers in the U.S. projected to reach 400
million?
10. A rental truck company charges a flat fee of $39.95 plus 59 cents per mile for a 26-foot
truck. Assume that there is no sales tax.
(a) What information above tells you that the relationship between the total cost and the
number of miles driven is linear?
(b) Determine the algebraic representation for this function.
(c) If a person’s bill was $90.69, how many miles must they have driven?
(d) Suppose there is a 6% sales tax. Determine a new algebraic representation for the
function relating total cost to miles driven, and use this function to answer part (c).
11. For each of the linear functions below, determine the vertical and horizontal intercepts.
(a)
(b)
(c)
12. Earlier this semester, we determined two ways of determining percentage changes.
Either you calculate
or you calculate
calculations are equivalent. In other words, show that
. Show that these two
.
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Section 7 Answers to Selected Homework Exercises
1. (a)
is a solution
2. (a)
x
0
1
2
3
4
-5
-1
3
7
11
Solution (Yes/No)
Yes
Yes
Yes
No
No
The table above confirms that
when x < 3.
3. (a)
(c)
(d)
(g)
(i)
(k)
4. (a)
(b)
(d)
(f)
(g)
(i) There are no solutions to this equation.
5. (a)
(b)
7. (a)
(b) In the year 2020, t = 12 years. So, P = 63.608 million people.
(c) Determining t when P = 60 million people yields t = 6.309 years. In other words, the
population is projected to reach 60 million people in the year 2014.
(d) Determining t when P = 112 million people yields t = 88.3 years. In other words,
the population is projected to reach 112 million people in the year 2008 + 88 =
2096.
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8. (a) In class, we let x represent the number of minutes used each month and y
represented the total monthly cost (in dollars). The linear function relating these two
variables was
.
(b)The equation that would need to be solved is
. The solution of
this equation is x = 340.5. Consequently, a person should use 340.5 minutes each
month if they want to spend exactly $50 per month.
(c) The inequality that would need to be solved is
. The solution is
. Consequently, a person can talk between 0-540.05 minutes each month
and not exceed $70.
9. (a) The population is projected to reach 7.5 billion people in the year 2016.
(b) The number of internet servers is projected to reach 400 million in the year 2009.
11. (a) Vertical intercept is y = -7; horizontal intercept is x = 7/4.