Transcript Introduction to Problem Solving

Equations and Inequalities
Copyright © Cengage Learning. All rights reserved.
2
Section
2.5
Introduction to Problem
Solving
Copyright © Cengage Learning. All rights reserved.
Objectives
1. Solve a number application using a linear
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equation in one variable.
22. Solve a geometry application using a linear
equation in one variable.
33. 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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1. Solve a number application using a
linear equation in one variable
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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
cont’d
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
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Example 1 – Plumbing
cont’d
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
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Example 1 – Plumbing
cont’d
3. Solve the equation
x + (x + 2) + 3x = 17
x + x + 3x = 17
5x + 2 = 17
5x = 15
x=3
(shortest)
3x = 9
x+2=5
(longest)
(middle)
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Example 1 – Plumbing
cont’d
4. Check the result
Because the sum of 3 feet, 5 feet, and 9 feet is 17 feet, the
solution checks.
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2. Solve a geometry application using a
linear equation in one variable
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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).
(a)
(b)
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.
Figure 2-8 (e)
1. What am I asked to find?
The unknown angle measure is designated as x degrees.
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Example 2 – Geometry
cont’d
2. Form an equation
x + 37 = 75
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Example 2 – Geometry
cont’d
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.
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3. Solve an investment application using a
linear equation in one variable
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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
cont’d
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
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Example – Investments
cont’d
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.
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Example – Investments
cont’d
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
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Example – Investments
cont’d
–3x = –13,500
x = 4,500
(investment @ 6%)
12000 – 4500 = 7500 (investment @ 9%)
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Example 8 – Investments
cont’d
4. Check the result
Interest on $4,000 @ 6% = $270
Interest on $7,500 @ 9% = $675
Total interest = $270 + $675 = $945
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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 = 412.50
Total income: 300 + 412.50
= 712.50
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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 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 = 350
x = 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 = 350
Total 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 – 0.01
Income from $8,000: x ∙ 8000
Income 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 – 200
8000x = 12000x – 320
320 = 4000x
x = 320/4000 = 8/100
= 0.08 (8%)
4. Check the answer
• Income from 8%: 8000 ∙ 0.08 = 640
Income from 7%: 12000 ∙ 0.07 = 840
640 is 200 less than 840
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