Copyright © Cengage Learning. All rights reserved.

Equations and Inequalities2

Copyright © Cengage Learning. All rights reserved.

Section 2.52.5

Introduction to Problem Solving

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Objectives

1. Solve a number application using a linear equation in one variable.

2. Solve a geometry application using a linear equation in one variable.

3. Solve an investment application using a linear equation in one variable.

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Problem Solving

We will use the following problem-solving strategy.

Problem Solving

1.What am I asked to find?Choose a variable to represent it.

2.Form an equation Relate the variable with all other unknowns in the problem

3.Solve the equation

4.Check the result

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Solve a number application using a linear equation in one variable

1.

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Example 1 – Plumbing

A plumber wants to cut a 17-foot pipe into three parts. (See Figure 2-7.) If the longest part is to be 3 times as long as the shortest part, and the middle-sized part is to be 2 feet longer than the shortest part, how long should each part be?

Figure 2-7

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Example 1 – Plumbing

1. What am I asked to find?

Length of the shortest part: x

Length of the longest part: 3x

Length of the middle part: x + 2

cont’d

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Example 1 – Plumbing

2. Form an Equation

The sum of the lengths of these three parts is equal to the total length of the pipe.

We can solve this equation.

x + (x + 2) + 3x = 17

cont’d

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Example 1 – Plumbing

3. Solve the equation

x + (x + 2) + 3x = 17

x + x + 3x = 17

5x + 2 = 17

5x = 15

x = 3 (shortest)

3x = 9 (longest) x + 2 = 5 (middle)

cont’d

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Example 1 – Plumbing

4. Check the result

Because the sum of 3 feet, 5 feet, and 9 feet is 17 feet, the solution checks.

cont’d

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Solve a geometry application using a linear equation in one variable

2.

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Problem Solving

The geometric figure shown in Figure 2-8(a) is an angle. Angles are measured in degrees.

The angle shown in Figure 2-8(b) measures 45 degrees (denoted as 45).

(b)(a)

Figure 2-8

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Problem Solving

If an angle measures 90, as in Figure 2-8(c), it is a right angle.

If an angle measures 180, as in Figure 2-8(d), it is a straight angle.

Adjacent angles are two angles that share a common side.

(c) (d)Figure 2-8

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Example 2 – Geometry

Refer to Figure 2-8(e) and find the value of x.

1.What am I asked to find?

The unknown angle measure is designated as x degrees.

Figure 2-8 (e)

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Example 2 – Geometry

2. Form an equation

x + 37 = 75

cont’d

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Example 2 – Geometry

3. Solve the equation

x + 37 = 75

x + 37 – 37 = 75 – 37

x = 38

4. Check the result

Since the sum of 38 and 37 is 75, the solution checks.

cont’d

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Solve an investment application using a linear equation in one variable

3.

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Example – Investments

A teacher invests part of $12,000 at 6% annual simple interest, and the rest at 9%. If the annual income from these investments was $945, how much did the teacher invest at each rate?

1.What am I asked to find?

We are asked to find the amount of money the teacher has invested in two different accounts.

Amount invested at 6%: x

Amount invested at 9%: 12000 - x

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Example – Investments

2. Form an equation

The interest i earned by an amount p invested at an annual rate r for t years is given by the formula i = prt.

In this example, t = 1 year. Hence, if x dollars were invested at 6%, the interest earned would be 0.06x dollars.

At 6%, interest = x(0.06)1At 9%, interest = (12000 – x)(0.09)1

Total interest = x(0.06)1 + (12000 – x)(0.09)1945 = x(0.06)1 + (12000 – x)(0.09)1

cont’d

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Example – Investments

The total interest earned in dollars can be expressed in two ways: as 945 and as the sum 0.06x + 0.09(12,000 – x).

We can form an equation as follows.

cont’d

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Example – Investments

3. Solve the equation

945 = x(0.06)1 + (12000 – x)(0.09)1

0.06x + 0.09(12,000 – x) = 945

6x + 9(12,000 – x) = 94,500

6x + 108,000 – 9x = 94,500

–3x + 108,000 = 94,500

cont’d

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Example – Investments

–3x = –13,500

x = 4,500 (investment @ 6%)

12000 – 4500 = 7500 (investment @ 9%)

cont’d

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Example 8 – Investments

4. Check the result

Interest on $4,000 @ 6% = $270Interest on $7,500 @ 9% = $675

Total interest = $270 + $675 = $945

cont’d

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Problem Solving: Investments

One investment pays 8% and another pays 11%. If equal amounts are invested in each, the combined interest income for 1 year is $712.50. How much was invested at each rate?

1. What am I asked to find?• Amount invested at 8%: x

Amount invested at 11%: x

2. Equation to relate known and unknown quantities• 0.08x + 0.11x = 712.50

3. Solve the equation• 0.19x = 712.50

x = 712.50 /0.19 = 71250 /19 = 3750

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Problem Solving: Investment (cont’)

4. Check the answer• Income from 8% investment: 0.08 ∙ 3750 = 300

Income from 11% investment: 0.11 ∙ 3750 = 412.50Total income: 300 + 412.50 = 712.50

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Problem Solving: Investment 2A college professor wants to supplement her retirement income with investment interest. If she invested $15,000 at 6% annual interest, how much more would she have to invest at 7% to achieve a goal of $1,250 per year in supplementary income?

1.What am I asked to find?•Amount to invest @ 7% interest : x

2.Form an equation•Income from 6% investment: 0.06 ∙ 15000 Income from 7% investment: 0.07x

0.06 ∙ 15000 + 0.07x = 1250

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Problem Solving: Investment 2

3. Solve the equation• 0.07x = 1250 – (0.06 ∙ 15000)

= 1250 – 900 = 350x = 350/0.07 = 35000/7 = 5000

4. Check the answer• Income from 6% investment: 0.06 ∙ 15000 = 900

Income from 7% investment: 0.07 ∙ 5000 = 350Total income: 900 + 350 =1250

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Problem Solving: Investment 3

The amount of annual interest earned by $8,000 invested at certain rate is $200 less than $12,000 would earn at 1% lower rate. A what rate is the $8,000 invested?

1.What am I asked to find?•Interest rate at which $8,000 is invested: x

2.Form an equation•Interest rate at which $12,000 is invested: x – 0.01Income from $8,000: x ∙ 8000Income from $12,000: (x – 0.01) ∙ 12000

x ∙ 8000 = (x – 0.01) ∙ 12000 - 200

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Problem Solving: Investment 3

3. Solve the equation• x ∙ 8000 = (x – 0.01) ∙ 12000 – 200

8000x = 12000x – 120 – 2008000x = 12000x – 320320 = 4000xx = 320/4000 = 8/100 = 0.08 (8%)

4. Check the answer• Income from 8%: 8000 ∙ 0.08 = 640

Income from 7%: 12000 ∙ 0.07 = 840640 is 200 less than 840