Download Solving Linear Inequalities in One Variable

Equations and Inequalities
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2
Section
2.5
Introduction to Problem
Solving
Copyright © Cengage Learning. All rights reserved.
Objectives
1. Solve a number application using a linear
1
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.
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 figures shown below is an angle.
Angles are measured in degrees.
The angle shown in Figure (b) measures 45 degrees
(denoted as 45).
(a)
(b)
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Problem Solving
If an angle measures 90, as in Figure (c), it is a right
angle.
If an angle measures 180, as in (d), it is a straight angle.
Adjacent angles are two angles that share a common
side.
(c)
(d)
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Example 2 – Geometry
Refer to Figure (e) and find the value of x.
(e)
Find the size angle x.
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
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
–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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Your Turn
Problem: A farmer owns a rectangular piece of land
whose length is 500 ft and its width is 40% of length. What
is the perimeter of the property?
1. What is to be found?
Let perimeter be p
length = 500; width = 0.40 · length
2. Form an equation
perimeter = 2 · length + 2 · width
p = 2(500) + 2(0.4)(500)
3. Solve the equation
p = 1000 + 400 = 1400
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