Download Chapter 3

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LESSON
3.1
Date ————————————
Study Guide
For use with pages 132–140
GOAL
Solve one-step equations using algebra.
Vocabulary
Inverse operations are two operations that undo each other, such as
addition and subtraction.
Equivalent equations are equations that have the same solution(s).
Properties of Equality
Addition Property of Equality Adding the same number to each side
of an equation produces an equivalent equation.
Subtraction Property of Equality Subtracting the same number from
each side of an equation produces an equivalent equation.
Division Property of Equality Dividing each side of an equation by
the same nonzero number produces an equivalent equation.
EXAMPLE 1
LESSON 3.1
Multiplication Property of Equality Multiplying each side of an
equation by the same nonzero number produces an equivalent equation.
Solve an equation using subtraction
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Solve x 1 11 5 15.
Solution
x 1 11 5 15
Write original equation.
x 1 11 2 11 5 15 2 11
Use subtraction property of equality:
Subtract 11 from each side.
x54
Simplify.
The solution is 4. Check by substituting 4 for x in the original equation.
x 1 11 5 15
4 1 11 5 15
15 5 15 ✓
CHECK
EXAMPLE 2
Write original equation.
Substitute 4 for x.
Solution checks.
Solve an equation using addition
Solve x 2 8 5 17.
Solution
Horizontal format
x 2 8 5 17
x 2 8 1 8 5 17 1 8
x 5 25
Vertical format
Write original equation.
x 2 8 5 17
18
Add 8 to each side.
Simplify.
x
18
5 25
The solution is 25.
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continued
For use with pages 132–140
Exercises for Examples 1 and 2
Solve the equation. Check your solution.
EXAMPLE 3
1. x 1 9 5 5
2.
y 1 2 5 25
3. 19 5 w 1 13
4. 8 5 z 2 11
5.
m2357
6. n 2 4 5 212
Solve an equation using division
Solve 7x 5 263.
LESSON 3.1
7x 5 263
7x
7
Write original equation.
263
7
}5}
Divide each side by 7.
x 5 29
EXAMPLE 4
Simplify.
Solve an equation using multiplication
x
Solve }
5 4.
12
x 5 48
EXAMPLE 5
Write original equation.
Multiply each side by 12.
Simplify.
Solve an equation by multiplying by a reciprocal
3
Solve }
x 5 6.
5
3
3
5
The coefficient of x is }5 . The reciprocal of }5 is }3.
3
5
}x 5 6
1 2
5 3
3 5
5
3
} }x 5 } (6)
x 5 10
Write original equation.
5
Multiply each side by the reciprocal, }3.
Simplify.
Exercises for Examples 3, 4, and 5
Solve the equation. Check your solution.
7. 29x 5 236
y
10. 18 5 }
22
10
Algebra 1
Chapter 3 Resource Book
8.
11.
7y 5 21
2
2}5 z 5 8
x
9. } 5 224
3
4
12. 16 5 } m
7
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
}54
12
x
12 p }
5 12 p 4
12
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LESSON
3.2
Date ————————————
Study Guide
For use with pages 141–146
GOAL
EXAMPLE 1
Solve two-step equations.
Solve a two-step equation
Solve 3x 2 7 5 8.
Solution
3x 2 7 5 8
Write original equation.
3x 2 7 1 7 5 8 1 7
Add 7 to each side.
3x 5 15
Simplify.
3x
3
Divide each side by 3.
15
3
}5}
x55
Simplify.
The solution is 5. Check by substituting 5 for x in the original equation.
CHECK
3x 2 7 5 8
Write original equation.
3(5) 2 7 5 8
Substitute 5 for x.
15 2 7 5 8
Multiply 3 and 5.
858✓
Simplify. Solution checks.
LESSON 3.2
Solve the equation. Check your solution.
EXAMPLE 2
x
1. } 2 3 5 5
4
2. 9y 1 2 5 29
3. 10 5 23z 2 8
m
4. 217 5 } 2 9
3
Solve a two-step equation by combining like terms
Solve 11x 2 9x 5 14.
Solution
11x 2 9x 5 14
2x 5 14
Combine like terms.
2x
2
Divide each side by 2.
14
2
}5}
x57
20
Algebra 1
Chapter 3 Resource Book
Write original equation.
Simplify.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Exercises for Example 1
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LESSON
3.2
Study Guide
Date ————————————
continued
For use with pages 141–146
Exercises for Example 2
Solve the equation. Check your solution.
EXAMPLE 3
5. 3x 1 11x 5 28
6. 7y 2 9y 5 12
7. 221 5 15w 2 12w
8. 36 5 4p 1 5p
Find an input of a function
The output of a function is 7 more than twice the input. Find the input
when the output is 13.
Solution
STEP 1
Write an equation for the function. Let x be the input and y be the output.
y 5 2x 1 7
STEP 2
Solve the equation for x when y 5 13.
y 5 2x 1 7
13 5 2x 1 7
13 2 7 5 2x 1 7 2 7
Write original equation.
Substitute 13 for y.
Subtract 7 from each side.
6 5 2x
Simplify.
}5}
6
2
Divide each side by 2.
35x
Simplify.
2x
2
LESSON 3.2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
y is 7 more than twice x.
An input of 3 produces an output of 13.
Exercises for Example 3
9. The output of a function is 8 less than 23 times the input. Find the input when
the output is 223.
10. The output of a function is 12 more than 5 times the input. Find the input when
the output is 28.
11. The output of a function is 7 more than 22 times the input. Find the input when
the output is 29.
12. The output of a function is 4 less than 9 times the input. Find the input when
the output is 68.
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LESSON
3.3
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Study Guide
For use with pages 1482153
GOAL
EXAMPLE 1
Solve multi-step equations.
Solve an equation by combining like terms
Solve 17x 2 11x 1 8 5 20.
Solution
17x 2 11x 1 8 5 20
Write original equation.
6x 1 8 5 20
Combine like terms.
6x 1 8 2 8 5 20 2 8
Subtract 8 from each side.
6x 5 12
Simplify.
6x
6
Divide each side by 6.
12
6
}5}
x52
Simplify.
Exercises for Example 1
Solve the equation. Check your solution.
1. 9x 2 13x 1 7 5 31
3. 15x 2 9 2 8x 5 12
4. 18 2 2x 2 4x 5 224
EXAMPLE 2
Solve an equation using the distributive property
Solve 4x 1 3(2x 2 1) 5 17.
Solution
METHOD 1 Show All Steps
4x 1 3(2x 2 1) 5 17
4x 1 6x 2 3 5 17
10x 2 3 5 17
LESSON 3.3
10x 2 3 1 3 5 17 1 3
32
10x 5 20
10x
10
20
10
}5}
x52
Algebra 1
Chapter 3 Resource Book
METHOD 2 Do Some Steps Mentally
4x 1 3(2x 2 1) 5 17
4x 1 6x 2 3 5 17
10x 2 3 5 17
10x 5 20
x52
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2. 13 2 5x 1 8x 5 22
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LESSON
3.3
Study Guide
Date ————————————
continued
For use with pages 1482153
Exercises for Example 2
Solve the equation. Check your solution.
5. 3(x 2 4) 1 4x 5 16
6. 9x 2 6(3x 2 3) 5 9
7. 22x 1 7(3x 21) 5 31
8. 5(2x 1 8) 2 6x 5 16
EXAMPLE 3
Multiply by a reciprocal to solve an equation
3
Solve }
(5x 2 4) 5 12.
4
Solution
3
} (5x 2 4) 5 12
4
4
3
3
4
4
3
} p } (5x 2 4) 5 } p 12
5x 2 4 5 16
x54
4
3
Multiply each side by }3, the reciprocal of }4 .
Simplify.
Subtract 4 from each side.
Simplify.
Exercises for Example 3
Solve the equation. Check your solution.
1
9. } (x 2 11) 5 9
2
3
10. 2} (2y 1 6) 5 15
2
5
11. 215 5 } (4z 2 1)
7
3
12. 36 5 2} (5m 1 12)
4
LESSON 3.3
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5x 5 20
Write original equation.
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LESSON
LESSON 3.4
3.4
Date ————————————
Study Guide
For use with pages 154 –160
GOAL
Solve equations with variables on both sides.
Vocabulary
An equation that is true for all values of the variable is an identity.
EXAMPLE 1
Solve an equation with variables on both sides
Solve 13 2 6x 5 3x 2 14.
Solution
13 2 6x 5 3x 2 14
Write original equation.
13 2 6x 1 6x 5 3x 2 14 1 6x
Add 6x to each side.
13 5 9x 2 14
Simplify.
27 5 9x
Add 14 to each side.
35x
Divide each side by 9.
The solution is 3. Check by substituting 3 for x in the original equation.
13 2 6x 5 3x 2 14
Write original equation.
13 2 6(3) 5 3(3) 2 14
Substitute 3 for x.
25 5 3(3) 2 14
Simplify left side.
25 5 25 ✓
Simplify right side. Solution checks.
Exercises for Example 1
Solve the equation. Check your solution.
1. 9a 5 7a 2 8
EXAMPLE 2
2.
17 2 8b 5 3b 2 5
3. 25c 1 6 5 9 2 4c
Solve an equation with grouping symbols
1
Solve 4x 2 7 5 }
(9x 2 15).
3
Solution
1
4x 2 7 5 }3 (9x 2 15)
Write original equation.
4x 2 7 5 3x 2 5
Distributive property
x 2 7 5 25
x52
The solution is 2.
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Subtract 3x from each side.
Add 7 to each side.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
CHECK
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LESSON
3.4
Study Guide
Date ————————————
continued
For use with pages 154 –160
LESSON 3.4
Exercises for Example 2
Solve the equation. Check your solution.
4. 2m 2 7 5 3(m 1 8)
1
5. } (15n 1 5) 5 8n 2 9
5
3
6. 7p 2 3 5 } (8p 2 12)
4
EXAMPLE 3
Identify the number of solutions of an equation
Solve the equation, if possible.
a. 4(3x 2 2) 5 2(6x 1 1)
b. 4(4x 2 5) 5 2(8x 2 10)
Solution
a. 4(3x 2 2) 5 2(6x 1 1)
12x 2 8 5 12x 1 2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
12x 5 12x 1 10
Write original equation.
Distributive property
Add 8 to each side.
The equation 12x 5 12x 1 10 is not true because the number 12x cannot be
equal to 10 more than itself. So, the equation has no solution. This can be
demonstrated by continuing to solve the equation.
12x 2 12x 5 12x 1 10 212x
0 5 10
Subtract 12x from each side.
Simplify.
The statement 0 5 10 is not true, so the equation has no solution.
b. 4(4x 2 5) 5 2(8x 2 10)
16x 2 20 5 16x 2 20
Write original equation.
Distributive property
Notice that the statement 16x 2 20 5 16x 2 20 is true for all values of x. So, the
equation is an identity.
Exercises for Example 3
Solve the equation, if possible.
7. 11x 1 7 5 10x 2 8
8. 5(3x 2 2) 5 3(5x 2 1)
1
9. } (6x 1 18) 5 3(x 1 3)
2
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LESSON
3.5
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Study Guide
For use with pages 162–167
GOAL
Find ratios and write and solve proportions.
Vocabulary
A ratio uses division to compare two quantities. You can write a ratio
of two quantities a and b, where b is not equal to 0, in three ways.
a to b
a:b
a
b
}
Each ratio is read “the ratio of a to b.” Ratios should be written in
simplest form.
A proportion is an equation that states that two ratios are equivalent.
Write a ratio
A shoe store sells 15 pairs of women’s shoes and 12 pairs of men’s
shoes. Find the specified ratio.
LESSON 3.5
EXAMPLE 1
a. Find the ratio of women’s shoe sales to men’s shoe sales.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b. Find the ratio of men’s shoe sales to total shoe sales.
Solution
women’s shoe sales
15
5
a. }} 5 } 5 }
12
4
men’s shoe sales
men’s shoe sales
12
12
4
b. }} 5 } 5 } 5 }
15 1 12
27
9
total shoe sales
EXAMPLE 2
Solve a proportion
7
x
Solve the proportion }
5}
.
9
Solution
7
x
}5}
9
54
7
x
54 p }9 5 54 p }
54
378
9
}5x
42 5 x
54
Write original proportion.
Multiply each side by 54.
Simplify.
Divide.
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continued
For use with pages 162–167
Exercises for Examples 1 and 2
A family has 75 VCR movies and 120 DVD movies. Find the specified
ratio.
1. The number of VCR movies to the number of DVD movies
2. The number of DVD movies to the entire movie collection
Solve the proportion. Check your solution.
LESSON 3.5
3
x
3. } 5 }
72
8
y
9
4. } 5 }
18
90
96
z
5. } 5 }
16
128
EXAMPLE 3
Solve a multi-step problem
A restaurant owner uses 3 cloves of garlic for every 5 pints of sauce. The restaurant uses
210 pints of sauce during the day. Find the number of cloves of garlic the restaurant
uses to make the sauce.
STEP 1
Write a proportion involving two ratios that compare the amount of garlic to
the pints of sauce.
3
5
cloves of garlic
pints of sauce
x
210
}5}
STEP 2
Solve the proportion.
3
5
x
210
}5}
3
Write original proportion.
x
210 p }5 5 210 p }
210
Multiply each side by 210.
}5x
630
5
Divide.
126 5 x
Simplify.
The restaurant uses 126 cloves of garlic to make the sauce.
Exercises for Example 3
6. A secretary can type 32 words in 60 seconds. How many words can the
secretary type in 150 seconds?
7. A secretary can type 48 words in 60 seconds. How many words can the
secretary type in 400 seconds?
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Solution
Name ———————————————————————
LESSON
3.6
Date ————————————
Study Guide
For use with pages 168–173
GOAL
Solve proportions using cross products.
Vocabulary
In a proportion, a cross product is the product of the numerator of one
ratio and the denominator of the other ratio.
Cross Products Property
The cross products of a proportion are equal.
a
b
c
d
If } 5 } where b Þ 0 and d Þ 0, then ad 5 bc.
A scale drawing is a two-dimensional drawing of an object in which
the dimensions of the drawing are in proportion to the dimensions of
the object.
A scale model is a three-dimensional model of an object in which the
dimensions of the model are in proportion to the dimensions of the
object.
The scale of a scale drawing or scale model relates the drawing’s or
model’s dimensions and the actual dimensions.
Solve a proportion using the cross products property
20
8
Solve }
5}
.
35
x
Solution
20
8
}5}
35
x
20 p x 5 8 p 35
20x 5 280
x 5 14
LESSON 3.6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
EXAMPLE 1
Write original proportion.
Cross products property
Simplify.
Divide each side by 20.
The solution is 14. Check your solution by substituting 14 for x in the original
proportion.
Exercises for Example 1
Solve the proportion. Check your solution.
15 126
1. } 5 }
x
210
2.
y18
21
y
9
}5}
24
28
3. } 5 }
z25
z
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continued
For use with pages 168–173
EXAMPLE 2
Write and solve a proportion
A bag of large breed dog food recommends feeding a dog 3 cups of food a day for
every 40 pounds of body weight. A dog weights 98 pounds. How much food should
the dog be eating each day?
Solution
STEP 1
Write a proportion involving two ratios that compare the amount of dog food
to the weight of the dog.
cups of food
3
x
}5}
40
98
weight of dog
STEP 2
Solve the proportion.
3
40
x
98
}5}
Write proportion.
3 p 98 5 40 p x
Cross products property
294 5 40x
Simplify.
7.35 5 x
Divide each side by 40.
A 98-pound dog should eat 7.35 cups of food each day.
Use the scale on a blueprint
LESSON 3.6
A blueprint of an office building has a scale of 2 inches:15 feet. A completed scale model
of the building is about 14.5 inches tall. Estimate the actual height of the office building.
Solution
STEP 1
Write a proportion to find the height x of the office building.
14.5
x
2
15
}5}
STEP 2
inches
feet
Solve the proportion.
2
15
14.5
x
}5}
2 p x 5 14.5 p 15
2x 5 217.5
x 5 108.75
Write proportion.
Cross products property
Simplify.
Divide each side by 2.
The height of the office building is about 108.75 feet.
Exercises for Examples 2 and 3
4. A car travels 135 miles on 4 gallons of gasoline. How many gallons of gasoline
will be used to travel 540 miles?
A blueprint has a scale of 3 cm : 5 m. Use the given measurement to
find the actual distance.
5. 4.5 cm
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6.
8.1 cm
7. 0.6 cm
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
EXAMPLE 3
Name ———————————————————————
LESSON
3.7
Date ————————————
Study Guide
For use with pages 176–181
GOAL
Solve percent problems.
Solving Percent Problems Using Proportions
You can represent “a is p percent of b” using the proportion
p
100
a
b
}5}
p
where a is a part of the base b and }
, or p%, is the percent.
100
The Percent Equation
You can represent “a is p percent of b” using the equation
a 5 p% p b
where a is a part of the base b and p% is the percent.
EXAMPLE 1
Find a percent using a proportion
What percent of 32 is 20?
Solution
p
a
100
b
p
20
}5}
32
100
}5}
Write proportion.
Substitute 20 for a and 32 for b.
2000 5 32x
Cross products property
62.5 5 x
Divide each side by 32.
.
20 is 62.5% of 32.
Exercises for Example 1
Use a proportion to answer the question.
1. What percent of 12 is 9?
2. What number is 46% of 150?
LESSON 3.7
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Write a proportion where 32 is the base and 20 is a part of the base.
3. 21 is 35% of what number?
4. What number is 72.5% of 240?
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continued
For use with pages 176–181
EXAMPLE 2
Find a percent using the percent equation
What percent of 25 is 12?
Solution
a 5 p% • b
12 5 p% • 25
0.48 5 p%
48% 5 p%
Write percent equation.
Substitute 12 for a and 25 for b.
Divide each side by 25.
Write decimal as percent.
12 is 48% of 25.
EXAMPLE 3
Find a part of a base using the percent equation
What number is 30% of 65?
Solution
a 5 p% p b
Write percent equation.
5 30% p 65
Substitute 30 for p and 65 for b.
5 0.30 p 65
Write percent as decimal.
5 19.5
Multiply.
EXAMPLE 4
Find a base using the percent equation
8 is 25% of what number?
Solution
a 5 p% p b
Write percent equation.
8 5 25% p b
Substitute 8 for a and 25 for p.
8 5 0.25 p b
Write percent as decimal.
32 5 b
Divide each side by 0.25.
LESSON 3.7
8 is 25% of 32.
Exercises for Examples 2, 3, and 4
Use the percent equation to answer the question.
5. What percent of 38 is 47.5?
6. What number is 46% of 130?
7. 12 is 40% of what number?
8. What percent of 250 is 85?
9. What number is 15% of 300?
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10. 17 is 25% of what number?
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
19.5 is 30% of 65.
Name ———————————————————————
LESSON
3.8
Date ————————————
Study Guide
For use with pages 184 –189
Write equations in function form and rewrite formulas.
Vocabulary
An equation in x and y is written in function form when the dependent
variable y is isolated on one side of the equation.
LESSON 3.8
GOAL
A literal equation is an equation that contains two or more variables.
EXAMPLE 1
Rewrite an equation in function form
Write 9x 2 4y 5 8 in function form.
Solution
To write an equation in function form, solve the equation for y.
9x 2 4y 5 8
Write original equation.
24y 5 8 2 9x
Subtract 9x from each side.
9
Divide each side by 24.
y 5 22 1 }4x
9
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The equation y 5 22 1 }4 x is written in function form.
EXAMPLE 2
Solve a literal equation
The formula for the volume of a rectangular prism is V 5 lwh. Solve the
formula for l.
Solution
V 5 lwh
V
wh
lwh
wh
Write original equation.
}5}
Assume w Þ 0 and h Þ 0. Divide each side by wh.
V
wh
Simplify.
}5l
V
wh
The rewritten equation is } 5 l.
Exercises for Examples 1 and 2
Write the equation in function form.
1. 7x 1 y 5 12
2.
3y 2 9x 5 21
3. 5y 2 2x 5 15
Solve the literal equation.
4. I 5 Prt for P
1
5. A 5 }(b1 1 b2)h for b2
2
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LESSON 3.8
3.8
Study Guide
Date ————————————
continued
For use with pages 184 –189
EXAMPLE 3
Solve and use a geometric formula
1
The area A of a triangle is given by the formula A 5 }
bh where b is the
2
base and h is the height.
12 in.
b
a. Solve the formula for the base b.
b. Use the rewritten formula to find the base of the triangle shown, which has an
area of 106.8 square inches.
Solution
a. Solve the formula for b.
1
A 5 }2bh
Write original formula.
2A 5 bh
Multiply each side by 2.
2A
}5b
h
Divide each side by h.
2A
h
b5}
2(106.8)
Write rewritten formula.
b5}
12
Substitute 106.8 for A and 12 for h.
b 5 17.8
Simplify.
The base of the triangle is 17.8 inches.
Exercises for Example 3
The surface area S of a sphere is given by the formula S 5 4πr 2 where r
is the radius of the sphere.
6. Solve the formula for r.
7. Use the rewritten formula from Exercise 6 to find r when S 5 314 square
meters. Use 3.14 for π.
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Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b. Substitute 106.8 for A and 12 for h in the rewritten formula.