Module 4: Problem Solving with Equations and Inequalities

Topics

1. Linear Equations in One Variable
2. An Introduction to Problem Solving
3. Linear Inequalities
4. Compound Inequalities
5. Absolute Value Equations
6. Absolute Value Inequalities

I. Linear Equations in One Variable

After completing this section, you should be able to:

• solve linear equations using properties of equality
• solve linear equations that can be simplified by combining like terms
• solve linear equations containing fractions
• identify equations that are identities and those that have no solution

A. Linear Equations in One Variable

A linear equation in one variable is a statement that two expressions are equal. For example,

3x + 5 = 8

is a linear equation in one variable. If the variable in the equation is replaced by a number and the resulting equation, or statement, is true, that number is called a solution of the equation.

For example, the number 1 is a solution of the equation 3x + 5 = 8 because:

3(1) + 5 = 8 is a true statement

But the number 2 is not a solution of this equation because:

3(2) + 5 = 8 is not a true statement

The solution set of an equation is the set of solutions of the equation. In the example above, the solution set is {1}.

To solve an equation we must find the solution set of the equation. Equations with the same solution set are called equivalent equations. For example:

• 3x + 5 = 8
• 3x = 3
• x = 1

are all equivalent equations because they all have the same solution set of {1}.

To solve an equation for a variable, we start with the given equation and write a series of simpler equivalent equations until we obtain an equation of the form:

x = number

To accomplish this, we use two important properties of equality, the addition and multiplication properties.

B. Addition and Multiplication Properties of Equality

In other words, the addition property of equality guarantees that when the same number is added to both sides of an equation, the result will be an equivalent equation.

Similarly, the multiplication property of equality guarantees that when the same number multiplies both sides of an equation, the result is an equivalent equation. Because subtraction can be defined in terms of addition and division can be defined in terms of multiplication, we can also subtract the same number from both sides of an equation, or divide both sides of an equation by the same nonzero number.

Example 1:	Solve for x:  2x + 5 = 9
Solution:	First, use the addition property of equality to subtract 5 from both sides. 2x + 5 = 9  Original equation. 2x + 5 –5 = 9 –5  Subtracted 5 from both sides. 2x = 4 Combined like terms. Now, use the multiplication property of equality to divide both sides by 2 to isolate the variable. =      Divided both sides by 2. x = 2      Simplified.
Check:	Check that 2 is the solution by substituting 2 for x in the original equation. 2x + 5 = 9      Original equation. 2(2) + 5 = 9    Replaced x with 2.                      4 + 5 = 9    Multiplied. 9 = 9    True.

C. Solving Linear Equations in One Variable

Often, before we can begin using the properties of equality, we must simplify each side of an equation by removing grouping symbols and combining like terms.

If an equation contains fractions, we first clear the equation of fractions by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions in the equation.

Example 3:	Solve for x:
Solution:	First, clear the fractions by multiplying both sides of the equation by 12, the LCD of the denominators. =   = Multiplied both sides by 12. 4y – 3y = 2 Distributed 12. y = 2 Combined like terms.

As a general guideline, the following steps may be used to solve a linear equation in one variable.

Lastly, we will look at linear equations that either have no solution or have more than one solution. These equations are called contradictions or identities, respectively.

• contradiction—an equation in one variable that has no solution
• identity—an equation in one variable that has all numbers as solutions

The equation 5 = 6 is a false statement no matter what value the variable x might have. Thus, the original equation has no solution. Its solution set is written either as:

{ } or Ø

Because 0 = 0 is a true statement for every value of x, all real numbers are solutions. The solution set is the set of all real numbers, which can be written:

{x | x is a real number}

II. An Introduction to Problem Solving

After completing this section, you should be able to:

• write algebraic expressions that can be simplified
• apply the steps for problem solving
• solve a formula for a specified variable
• use formulas to solve problems

A. Writing Algebraic Expressions from Word Statements

In module 1, we practiced translating phrases into algebraic expressions. In this section, we will practice writing algebraic expressions and then simplifying them.

The first example involves consecutive integers. Consecutive integers are integers that follow one another in order. For example, 6, 7, and 8 are three consecutive integers. We will also study consecutive even integers and consecutive odd integers. For example, 10, 12, and 14 are consecutive even integers, whereas 7, 9, and 11 are consecutive odd integers.

Using variables, here's how we represent consecutive integers:

You may wonder why we add 2 to the first consecutive odd integer to get the second consecutive odd integer. An example will make the reason clear. Consider the odd integer 7. The next odd integer is 9, which is 2 units more than the first integer. Thus, if the first consecutive odd integer is represented by x, the second will be 2 units more, or x + 2.

B. General Strategy for Problem Solving

The following problem-solving strategy will be used throughout this course.

Now let's see how to apply these rules.

Watch this animation explaining how to apply the steps for problem solving.

Example 4:	Suppose that a computer store just announced an 8% decrease in the price of a particular computer. If the computer sells for $2,162 after the decrease, find the original price of the computer.
Solution:	1. UNDERSTAND the problem. Read and reread the problem. Recall that a percent decrease means a percent of the original price. Let's guess that the original price of the computer is $2,500. The amount of decrease is then 8% of $2,500: (0.08)($2,500) = $200 Therefore, the new price of the computer is the original price minus the decrease: $2,500 – $200 = $2,300 Our guess is incorrect, but we now have an idea of how to model this problem. In our model, we will: let x = the original price of the computer 2. TRANSLATE the problem into an equation. To do this, we use the fact that the sum of the numbers is 72. First let's write this relationship in words and then translate to an equation. In words: 3. SOLVE the equation. x – 0.08x = 2,162 Original equation. Remember that x is the same as 1x. 0.92x = 2,162 Combined like terms. x = Divided both sides by 0.92. x = 2,350 Simplified. 4. INTERPRET the results. Here, we check our work and state the solution. Check your work: If the original price of the computer was $2,350, the new price is $2,162, as shown in the following calculation: $2,350 – (0.08)($2,350) = $2,350 – $188 = $2,162 State the solution: The original price of the computer was $2,350.

We've seen some problem-solving approaches to solving linear equations. Now let's look at problem solving using formulas.

C. Formulas and Problem Solving

A formula is an equation that describes a known relationship among quantities such as time, area, or volume. Some examples of formulas are given below.

Let's consider the last formula. Note that this formula is solved for the variable V, that is, V is by itself on one side of the equation and there are no V's on the other side. Sometimes, however, it is useful to solve a formula for different variable in the equation, that is, we want to have a different variable isolated on one side of the equation.

Example 5: Solve V = lwh for h.

Solution:

To isolate h on one side of the equation, divide both sides of the equation by lw. V = lwh Original formula. = Divided both sides by lw. = h Simplified.

The following steps may be used to solve formulas or equations for a specified variable.

Now let's see how to apply these rules.

Watch this animation explaining how to solve equations for a specified variable.

Example 6: Solve A = (B + b)h for b.

Solution:

Because this formula contains fractions, we begin by multiplying both sides of the equation by the LCD, which is 2. A = (B + b)h Original problem. 2 · A = 2 · (B + b)h Multiplied both sides by 2. 2A = (B + b)h Simplified. 2A = Bh + bh Distributed h. 2A – Bh = bh Isolated the term containing b. = Divided both sides by h. = b Simplified.

Next, we will solve problems that can be modeled by known formulas. We use the same problem-solving steps that were given in the previous section.

Now let's see how to apply these steps.

Watch this animation explaining how to use formulas to solve problems.

Example 7:	Karen just received an inheritance of $10,000 and plans to place all the money in a savings account that pays 5% interest compounded quarterly to help her son go to college in 3 years. How much money will be in the account in 3 years?
Solution:	1. UNDERSTAND. Read and reread the problem. The appropriate formula needed to solve this problem is the compound interest formula: Be sure you understand the meaning of each of the variables in this formula: A = amount in the account after t years P = principal (or starting amount) invested t = time in years r = annual rate of interest (expressed as a decimal) n = number of times compounded per year 2. TRANSLATE. Use the compound interest formula and let: P = $10,000 r = 5% = 0.05 t = 3 years n = 4 (n = 4 because the account is compounded quarterly, or 4 times per year.) 3. SOLVE the equation. Original formula. Substituted values for variables. A = 10,000(1.0125)12 Multiplied 4 · 3 in the exponent. A ≈ 10,000(1.160754518) Approximately (1.0125)12. A ≈ $11,607.55 Multiplied and rounded to two decimals. 4. INTERPRET the results. Here, we check our work and state the solution. Check your work: Repeat your calculations to make sure that no error was made. Notice that $11,607.55 is a reasonable amount to have in an account after 3 years. State the solution: In 3 years, the account will contain $11,607.55.

III. Linear Inequalities

After completing this section, you should be able to:

• use interval notation
• solve linear inequalities using the addition property of inequality
• solve linear inequalities using the multiplication property of inequality
• solve problems that can be modeled by linear inequalities

A. Linear Inequalities in One Variable

A linear inequality is similar to an equation except that the equality symbol is replaced with an inequality symbol, such as <, >, ≤, or ≥.

A solution of an inequality is a value of the variable that makes the inequality a true statement. The solution set of an inequality is the set of all solutions.

For example, the solution set of the inequality:

x > 2

contains all numbers greater than 2, and can be written in set notation as:

{x | x > 2}

In these modules, we will use interval notation to write the solution set. With this notation, the graph of:

{x | x > 2}

looks like:

and can be represented in interval notation as (2, ∞). Notice that the interval does not include the number 2. The graph of:

{x | x ≥ 2}

looks like:

and can be represented in interval notation as [2, ∞).

The following table shows three equivalent ways to describe an interval: in set notation, as a graph, and in interval notation.

B. Addition Property of Inequalities

To solve a linear inequality, we use a process similar to the one used to solve a linear equation. We use properties of inequalities to write successive equivalent inequalities until the variable is isolated.

In other words, the same number may be added or subtracted from both sides of an inequality without changing the solution to the inequality.

C. Multiplication Property of Inequality

To understand the multiplication property of inequality, let's start with the true statement:

      –3 < 7

and multiply both sides by a positive number.

       –3 < 7
  –3(2) < 7(2)      Multiplied by 2.
      –6 < 14        True.

The statement remains true when we multiply by 2. Next, let's multiply both sides of –3 < 7 by a negative number.

      –3 < 7
–3(–2) < 7(–2)     Multiplied by –2.
       6 < –14        False.

The statement is now false. However, if the direction of the inequality symbol is reversed, the statement will be true.

        6 > –14       True.

These examples lead to the multiplication property of inequality.

In other words, when we multiply both sides of an inequality by the same positive real number, the result is an equivalent inequality. When we multiply both sides of an inequality by the same negative real number, however, we must reverse the sign of the inequality symbol.

Example 2:	Solve and graph the solution set to the following inequality:x ≤
Solution:	x ≤    4 · x ≤ 4 ·   Multiplied both sides by 4.         x ≤         Simplified. The solution set is , which in interval notation is . The graph of the solution set is:

D. Solving a Linear Inequality in One Variable

To solve linear inequalities in general, follow the steps below.

Example 3: Solve –(x – 3) + 2 ≤ 3(2x – 5) + x

Solution:

–(x – 3) + 2 ≤ 3(2x – 5) + x Original equation. –x + 3 + 2 ≤ 6x – 15 + x Applied the distributive property. 5 –x ≤ 7x – 15 Combined like terms. 5 – x + x ≤ 7x + x – 15 Added x to both sides. 5 ≤ 8x – 15 Combined like terms. 5 + 15 ≤ 8x – 15 + 15 Added 15 to both sides. 20 ≤ 8x Combined like terms. ≤ Divided both sides by 8. ≤ x   or   x ≥ Simplified. Helpful Hint Don't forget that ≤ x is the same as x ≥ . The solution set written in interval notation is . Its graph is:

We conclude this section by looking at application problems that can be represented by an inequality rather than an equation. We look for words such as "at least," "at most," "between," "no more than," and "no less than." In solving applications involving linear inequalities, we use the same procedure we used when we solved applications involving linear equations.

Example 4:	Calculating Income with Commission
	A salesperson earns $600 per month plus a commission of 20% of his total sales. Find the minimum amount of sales needed to receive a total income of at least $1,500 per month.
Solution:	1. UNDERSTAND. Read and reread the problem. Let x = amount of sales. 2. TRANSLATE. As stated in the beginning of this example, we want the income greater than or equal to $1,500. To write an inequality, notice that the sales income consists of $600 plus a commission (20% of sales).
	3. SOLVE the inequality for x.   600 + 0.20x ≥ 1,500 600 + 0.20x – 600 ≥ 1,500 – 600 Subtracted 600 from both sides. 0.20 ≥ 900 Simplified. ≥ Divided both sides by 0.20. x ≥ 4,500 Simplified. 4. INTERPRET. Here, we check our work and state the solution. Check your work: The income for sales of $4,500 is: 600 + 0.20(4,500), or $1,500 Therefore, if sales are greater than or equal to $4,500, then the income is greater than or equal to $1,500. State the solution: The minimum amount of sales needed for the salesperson to earn at least $1,500 per month is $4,500 per month.

Now that we've solved simple linear inequalities (linear inequalities with one inequality symbol in the expression), we are now ready to discuss compound linear inequalities.

IV. Compound Inequalities

After completing this section, you should be able to:

• find the intersection of two sets
• find the union of two sets
• solve the compound inequalities containing and
• solve compound inequalities containing or

A. Intersection of Two Sets

B. Union of Two Sets

To see animations illustrating these operations, click on the buttons below the picture of sets A and B. The results of the operations are shaded in blue.

Figure 4.1
Intersection and Union

C. Solving Compound Inequalities

Two inequalities joined by the words and or or are called compound inequalities.
For example:

The solution set of a compound inequality formed by the word and is the intersection of the solutions sets of the two inequalities. A value is a solution of the compound inequality if it is a solution of both inequalities.

For example, the solution set of the compound inequality x ≤ 5 and x ≥ 3 contains all values of x that make the inequality x ≤ 5 and the inequality x ≥ 3 true statements.

The graphs below illustrate the solution sets of each inequality, as well as the solution set of the compound inequality.

Set Notation	Graph	Interval Notation
{x | x ≤ 5}		(–∞, 5]
{x | x ≥ 3}		[3, ∞)
{x | 3 ≤ x ≤ 5}		[3, 5]

We can write the solution set of the compound inequality in interval notation as [3, 5].

We can also write a compound inequality joined by the word and in compact form, such as:

2 < 4 –x < 7

To solve this compound inequality, we isolate x in the "middle part." Because a compound inequality is really two inequalities in one statement, we must perform the same operations on all three parts of the inequality.

In some cases, the intersection of the solution sets may be empty, and the compound inequality will have no solution.

There is no number that is greater than or equal to 0 and less than or equal to –2. Therefore, the solution set is ∅ (or empty set).

The solution set of a compound inequality formed by the word or is the union of the solution sets of the two inequalities. A value is a solution of the compound inequality if it is a solution of either inequality.

For example, the solution set of the compound inequality x ≤ 1 or x ≥ 3 contains all numbers that make the inequality x ≤ 1 a true statement or the inequality x ≥ 3 a true statement.

The graphs below illustrate the solution sets of each inequality as well as the solution set of the compound inequality.

We can write the solution set of the compound inequality in interval notation as:

(–∞, 1] ∪ [3, ∞)

Example 6: Solve 5x – 3 ≤ 10  or  x + 1 ≥ 5.

In some cases, the solution sets of each inequality overlap, and in such cases, the union of the solution sets will be the set of all real numbers. In this particular example, there is no overlap because there is no common portion of the intervals as you can see in the graphs above.

V. Absolute Value Equations

After completing this section, you should be able to:

• solve absolute value equations

A. Solving Equations of the Form |x|= a

In module 1, we defined the absolute value of a number as its distance from 0 on a number line. In this section, we concentrate on solving equations containing the absolute value of a variable or variable expression. Examples of absolute value equations are:

• |x|= 3
• –5 = |2x + 7|
• |x – 5|= |3x + 1|

For the absolute value equation |x|= 3, the solution set contains all numbers whose distance from 0 is 3 units. Two numbers are 3 units away from 0 on the number line: 3 and –3, as shown below:

Thus, the solution set of the equation |x|= 3 is {3, –3}.

The next example illustrates a special case for absolute value equations: the absolute value, when isolated, is equal to a negative number.

B. Solving Equations of the Form |x|= |y|

Two absolute value expressions are equal if the expressions inside the absolute value bars are equal to each other or are opposites of each other.

Example 3:	Solve |3x + 2|= |5x – 8|.
Solution:	This equation is true if the expressions inside the absolute value bars are equal to each other or are opposites of each other. Solve each equation: 3x + 2 = 5x –8 or 3x + 2 = –(5x – 8) –2x + 2 = –8 or 3x + 2 = –5x + 8 –2x = –10 or 8x + 2 = 8 x = 5 or 8x = 6 x = 5 or x =
	The solutions are and 5. We can check our answers by substituting 5 for x and then for x in the original equation: |3x + 2|= |5x – 8| We will leave checking these solutions for you to do.

Note: If we had solved only for the expressions inside the absolute value bars being equal to each other, we would have omitted the second solution to the equation, namely, the solution that results from setting the left side equal to the opposite of the right side. It is important to remember when solving absolute value expressions that you must solve two equations.

VI. Absolute Value Inequalities

After completing this section, you should be able to do the following:

• solve absolute value equations of the form |x| < a
• solve absolute value equations of the form |x| > a

A. Solving Absolute Value Inequalities of the Form |x| < a

The solution set of an absolute value inequality such as |x| < 2 contains all numbers whose distance from 0 is less than 2 units, as shown below.

We graph this solution set by indicating rounded parentheses at the endpoints (because the endpoints are not included) and drawing a line to represent the space between the endpoints.

The solution set is:

{x | –2 < x < 2}

or, in interval notation:

(–2, 2)

Below, we give the rule for solving absolute value inequalities of the form |x| < a.

Now let's see how to apply this rule.

Watch this animation explaining solve inequalities of the form |x| < a.

B. Solving Absolute Value Inequalities of the Form
 |x| > a

The solution set of an absolute value inequality such as |x| ≥ 3 contains all numbers whose distance from 0 is 3 or more units. This distance can be in either the positive or in the negative direction. The graph of the solution set contains the number 3 and all points to the right of 3 on the number line, and –3 and all points to the left of -3 on the number line, as shown below.

This graph represents the collection of points that begin 3 units from 0 (at –3 and 3) and continue away from 0 (hence having a distance of 3 or more from 0 in both directions).

The solution set is:

{x | x ≤ –3 or x ≥ 3}

or, in interval notation:

(–∞, -3] ∪ [3, ∞)

Below, we give the rule for solving absolute value inequalities of the form |x| > a.

We have concluded our discussion of solving absolute value inequality expressions. Now try the problems in the self-assessment section of this module to see how well you understand the concepts that you've just covered. The self-assessment problems are not graded, but are a useful tool to determine which concepts you've mastered and which concepts you may wish to review before moving forward.

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