Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2

Copyright © Cengage Learning. All rights reserved. Section 2.5 Introduction to Problem Solving

3 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

4 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

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

6 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

7 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

8 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

9 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

10 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

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

12 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

13 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

14 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)

15 Example 2 – Geometry 2.Form an equation x + 37 = 75 cont’d

16 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

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

18 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%: x

19 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)1 At 9%, interest = (12000 – x)(0.09)1 Total interest = x(0.06)1 + (12000 – x)(0.09)1 945 = x(0.06)1 + (12000 – x)(0.09)1 cont’d

20 Example – Investments The total interest earned in dollars can be expressed in two ways: as 945 and as the sum 0.06x (12,000 – x). We can form an equation as follows. cont’d

21 Example – Investments 3.Solve the equation 945 = x(0.06)1 + (12000 – x)(0.09)1 0.06x (12,000 – x) = 945 6x + 9(12,000 – x) = 94,500 6x + 108,000 – 9x = 94,500 –3x + 108,000 = 94,500 cont’d

22 Example – Investments –3x = –13,500 x = 4,500 6%) – 4500 = %) cont’d

23 Example 8 – Investments 4.Check the result Interest on 6% = $270 Interest on 9% = $675 Total interest = $270 + $675 = $945 cont’d

24 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 $ 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 x = Solve the equation 0.19x = x = /0.19 = /19 = 3750

25 Problem Solving: Investment (cont’) 4.Check the answer Income from 8% investment: 0.08 ∙ 3750 = 300 Income from 11% investment: 0.11 ∙ 3750 = Total income: =

26 Problem Solving: Investment 2 A 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 7% interest : x 2.Form an equation Income from 6% investment: 0.06 ∙ Income from 7% investment: 0.07x 0.06 ∙ x = 1250

27 Problem Solving: Investment 2 3.Solve the equation 0.07x = 1250 – (0.06 ∙ 15000) = 1250 – 900 = 350 x = 350/0.07 = 35000/7 = Check the answer Income from 6% investment: 0.06 ∙ = 900 Income from 7% investment: 0.07 ∙ 5000 = 350 Total income: =1250

28 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 – 0.01 Income from $8,000: x ∙ 8000 Income from $12,000: (x – 0.01) ∙ x ∙ 8000 = (x – 0.01) ∙

29 Problem Solving: Investment 3 3.Solve the equation x ∙ 8000 = (x – 0.01) ∙ – x = 12000x – 120 – x = 12000x – = 4000x x = 320/4000 = 8/100 = 0.08 (8%) 4.Check the answer Income from 8%: 8000 ∙ 0.08 = 640 Income from 7%: ∙ 0.07 = is 200 less than 840