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 . Is this equation true? Both the left and right sides equal 15, so it appears to be true. 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)

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(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 can be simplified down to 2v. An alternate way of thinking about this is as follows: 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)

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(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 Solution (Yes/No)

9 -6 No 0 0 Yes

(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 Solution (Yes/No)

3 Yes 10 No

t = 20 8 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 . This is because .

If our work above is correct, 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 Solution (Yes/No) 0 1 2 3 4

(b) ; solution:

t Solution (Yes/No) -1 0 1 2 3

3. Simplify each of the following expressions.

(a) (b) (c)

(d)

(e)

(f)

(g)

(h)

(i) (j)

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(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 . Show that these two

calculations are equivalent. In other words, show that .

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Section 7 Answers to Selected Homework Exercises

1. (a) is a solution

2. (a) x Solution (Yes/No) 0 -5 Yes 1 -1 Yes 2 3 Yes 3 7 No 4 11 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.