Download Solving Two-Step Equations

3.1
Solving Two-Step Equations
Goal: Solve two-step equations.
Example 1
Using Subtraction and Division to Solve
Solve 4x 9 7. Check your solution.
4x 9 7
Notice in Example 1
that you isolate x by
working backward.
First you subtract
from each side and
then you divide.
4x 9 Write original equation.
7 Subtract
4x Simplify.
4x
16
Divide each side by
x
.
4x 9 7
(
4
.
Simplify.
Answer: The solution is
Check:
from each side.
) 9 7
7
Write original equation.
Substitute for x.
.
Checkpoint Solve the equation. Check your solution.
1. 3x 8 26
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2. 21 4x 7
Chapter 3 • Pre-Algebra Notetaking Guide
43
3.1
Solving Two-Step Equations
Goal: Solve two-step equations.
Example 1
Using Subtraction and Division to Solve
Solve 4x 9 7. Check your solution.
4x 9 7
Notice in Example 1
that you isolate x by
working backward.
First you subtract
from each side and
then you divide.
Write original equation.
4x 9 9 7 9
4x 16
4x
16
4
4
x 4
Subtract 9 from each side.
Simplify.
Divide each side by 4 .
Simplify.
Answer: The solution is 4 .
4x 9 7
Check:
(
4 4
) 9 7
7
7
Write original equation.
Substitute for x.
Solution checks .
Checkpoint Solve the equation. Check your solution.
1. 3x 8 26
6
Copyright © Holt McDougal. All rights reserved.
2. 21 4x 7
7
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43
Using Addition and Multiplication to Solve
Example 2
x
3
Solve 4 1. Check your solution.
x
4 1
3
x
4 3
Write original equation.
1 x
3
x
3
Add
Simplify.
( )
Multiply each side by
x
.
Simplify.
Answer: The solution is
Check:
to each side.
x
4 1
3
4 1
3
.
Write original equation.
Substitute for x.
1
.
Checkpoint Solve the equation. Check your solution.
x
4
3. 7 2
44
Chapter 3 • Pre-Algebra Notetaking Guide
b
5
4. 8 3
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Using Addition and Multiplication to Solve
Example 2
x
3
Solve 4 1. Check your solution.
x
4 1
3
Write original equation.
x
4 4 1 4
3
x
3
3
x
3 3 3
3
( )
x 9
Add 4 to each side.
Simplify.
Multiply each side by 3 .
Simplify.
Answer: The solution is 9 .
Check:
x
4 1
3
9
4 1
3
1
1
Write original equation.
Substitute for x.
Solution checks .
Checkpoint Solve the equation. Check your solution.
x
4
b
5
3. 7 2
4. 8 3
36
44
Chapter 3 • Pre-Algebra Notetaking Guide
55
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Solving an Equation with Negative Coefficients
Example 3
Solve 2 3x 17. Check your solution.
2 3x 17
2 3x Write original equation.
17 Subtract
3x Simplify.
3x
15
Divide each side by
x
.
2 3x 17
23
(
.
Simplify.
Answer: The solution is
Check:
from each side.
) 17
17
Write original equation.
Substitute for x.
.
Checkpoint Solve the equation. Check your solution.
5. 3 2y 19
Copyright © Holt McDougal. All rights reserved.
6. 5 4 m
Chapter 3 • Pre-Algebra Notetaking Guide
45
Example 3
Solving an Equation with Negative Coefficients
Solve 2 3x 17. Check your solution.
2 3x 17
Write original equation.
2 3x 2 17 2
3x 15
Subtract 2 from each side.
Simplify.
3x
15
3
3
x 5
Divide each side by 3 .
Simplify.
Answer: The solution is 5 .
Check:
2 3x 17
(
2 3 5
17
) 17
17
Write original equation.
Substitute for x.
Solution checks .
Checkpoint Solve the equation. Check your solution.
5. 3 2y 19
8
Copyright © Holt McDougal. All rights reserved.
6. 5 4 m
9
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45
3.2
Solving Equations Having
Like Terms and Parentheses
Goal: Solve equations using the distributive property.
Writing and Solving an Equation
Example 1
Baseball Game A group of five friends are going to a baseball
game. Tickets for the game cost $12 each, or $60 for the group.
The group also wants to eat at the game. Hot dogs cost $2.75 each
and bottled water costs $1.25 each. The group has a total budget
of $85. If the group buys the same number of hot dogs and bottles
of water, how many can they afford to buy?
Solution
Let n represent the number of hot dogs and the number of bottles
of water. Then 2.75n represents the cost of n hot dogs and 1.25n
represents the cost of n bottles of water. Write a verbal model.
Substitute.
Combine like terms.
Subtract
each side.
Simplify.
Divide each side
by
.
n
from
Simplify.
Answer: The answer must be a whole number. Round down so the
budget is not exceeded. The group can afford to buy
hot dogs
and
bottles of water.
46
Chapter 3 • Pre-Algebra Notetaking Guide
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3.2
Solving Equations Having
Like Terms and Parentheses
Goal: Solve equations using the distributive property.
Example 1
Writing and Solving an Equation
Baseball Game A group of five friends are going to a baseball
game. Tickets for the game cost $12 each, or $60 for the group.
The group also wants to eat at the game. Hot dogs cost $2.75 each
and bottled water costs $1.25 each. The group has a total budget
of $85. If the group buys the same number of hot dogs and bottles
of water, how many can they afford to buy?
Solution
Let n represent the number of hot dogs and the number of bottles
of water. Then 2.75n represents the cost of n hot dogs and 1.25n
represents the cost of n bottles of water. Write a verbal model.
Cost of
Cost of
Total
Cost of
tickets budget
bottled water
hot dogs
2.75n 1.25n 60 85
Substitute.
4n 60 85
Combine like terms.
4n 60 60 85 60
Subtract 60 from
each side.
Simplify.
4n 25
4n
4
25
4
n
1
6 4
Divide each side
by 4 .
Simplify.
Answer: The answer must be a whole number. Round down so the
budget is not exceeded. The group can afford to buy 6 hot dogs
and 6 bottles of water.
46
Chapter 3 • Pre-Algebra Notetaking Guide
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Example 2
Solving Equations Using the Distributive Property
Solve the equation.
a. 24 6(2 x)
Solution
a.
24 b.
b. 2(7 4x) 10
24 6(2 x)
Write original equation.
24 Distributive property
from each side.
Simplify.
Divide each side by
x
Simplify.
2(7 4x) 10
.
Write original equation.
10
Distributive property
10 Add
Simplify.
Divide each side by
x
Copyright © Holt McDougal. All rights reserved.
Subtract
to each side.
.
Simplify.
Chapter 3 • Pre-Algebra Notetaking Guide
47
Solving Equations Using the Distributive Property
Example 2
Solve the equation.
a. 24 6(2 x)
Solution
a.
b. 2(7 4x) 10
24 6(2 x)
Write original equation.
24 12 6x
Distributive property
24 12 12 6x 12
36 6x
36
6
Simplify.
6x
Divide each side by 6 .
6
6 x
b.
Simplify.
2(7 4x) 10
14 8x 10
14 8x 14 10 14
8x 24
8x
Distributive property
Add 14 to each side.
Simplify.
Divide each side by 8 .
8
x 3
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Write original equation.
24
8
Subtract 12 from each side.
Simplify.
Chapter 3 • Pre-Algebra Notetaking Guide
47
Example 3
Combining Like Terms After Distributing
Solve 6x 4(x 1) 14.
6x 4(x 1) 14
6x
Write original equation.
14
Distributive property
14
Combine like terms.
14 Subtract
Simplify.
Divide each side by
x
from each side.
.
Simplify.
Checkpoint Solve the equation. Check your solution.
48
1. 20 5(3 x)
2. 4y 14 3y 28
3. 3(6 2x) 12
4. 5x 2(x 3) 30
Chapter 3 • Pre-Algebra Notetaking Guide
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Example 3
Combining Like Terms After Distributing
Solve 6x 4(x 1) 14.
6x 4(x 1) 14
Write original equation.
6x 4x 4 14
Distributive property
2x 4 14
Combine like terms.
2x 4 4 14 4
2x 10
2x
Subtract 4 from each side.
Simplify.
10
2
Divide each side by 2 .
2
x 5
Simplify.
Checkpoint Solve the equation. Check your solution.
1. 20 5(3 x)
7
3. 3(6 2x) 12
5
48
Chapter 3 • Pre-Algebra Notetaking Guide
2. 4y 14 3y 28
6
4. 5x 2(x 3) 30
8
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3.3
Solving Equations with
Variables on Both Sides
Goal: Solve equations with variables on both sides.
Solving an Equation with the Variable on Both Sides
Example 1
Solve 5n 7 9n 21.
5n 7 9n 21
5n 7 Write original equation.
9n 21 7 7 Subtract
21
Simplify.
21 Subtract
from each side.
from each side.
Simplify.
Divide each side by
n
Answer: The solution is
.
Simplify.
.
An Equation with No Solution
Example 2
Solve 3(2x 1) 6x.
3(2x 1) 6x
Write original equation.
6x
Distributive property
Notice that this statement
true because the number 6x
.
.
The equation has
As a check, you can continue solving the equation.
The statement
6x
Subtract
Simplify.
from each side.
true, so the equation has
.
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Chapter 3 • Pre-Algebra Notetaking Guide
49
3.3
Solving Equations with
Variables on Both Sides
Goal: Solve equations with variables on both sides.
Example 1
Solving an Equation with the Variable on Both Sides
Solve 5n 7 9n 21.
5n 7 9n 21
5n 7 5n 9n 21 5n
7 4n 21
7 21 4n 21 21
28 4n
Write original equation.
Subtract 5n from each side.
Simplify.
Subtract 21 from each side.
Simplify.
28
4n
4
4
Divide each side by 4 .
7 n
Simplify.
Answer: The solution is 7 .
Example 2
An Equation with No Solution
Solve 3(2x 1) 6x.
3(2x 1) 6x
Write original equation.
6x 3 6x
Distributive property
Notice that this statement is not true because the number 6x
cannot be equal to 3 more than itself . The equation has no
solution . As a check, you can continue solving the equation.
6x 3 6x 6x 6x
3 0
The statement 3 0
Subtract 6x from each side.
Simplify.
is not true, so the equation has
no solution .
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49
Example 3
Solving an Equation with All Numbers as Solutions
Solve 4(x 2) 4x 8.
4(x 2) 4x 8
4x 8
Write original equation.
Distributive property
Notice that for all values of x, the statement
4x 8 is
. The equation has
.
Checkpoint Solve the equation. Check your solution.
50
1. 3n 6 5n 20
2. 12 x 4(3x 1)
3. 3(2n 4) 2(3n 6)
4. 2x 7 2x 13
Chapter 3 • Pre-Algebra Notetaking Guide
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Example 3
Solving an Equation with All Numbers as Solutions
Solve 4(x 2) 4x 8.
4(x 2) 4x 8
4x 8 4x 8
Write original equation.
Distributive property
Notice that for all values of x, the statement 4x 8 4x 8 is
true . The equation has every number as a solution .
Checkpoint Solve the equation. Check your solution.
1. 3n 6 5n 20
13
3. 3(2n 4) 2(3n 6)
all numbers
50
Chapter 3 • Pre-Algebra Notetaking Guide
2. 12 x 4(3x 1)
no solution
4. 2x 7 2x 13
5
Copyright © Holt McDougal. All rights reserved.
Solving an Equation to Find a Perimeter
Example 4
Geometry Find the perimeter of
the square.
x6
3x
Solution
1. A square has four sides of equal length. Write an equation and
solve for x.
Write equation.
Subtract
from each side.
Simplify.
Divide each side by
x
Simplify.
2. Find the length of one side by substituting
expression.
( )
3x 3
for x in either
Substitute for x and multiply.
3. To find the perimeter, multiply the length of one side by
p
.
.
Answer: The perimeter of the square is
units.
Checkpoint Find the perimeter of the square.
5.
3x 8
5x
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51
Example 4
Solving an Equation to Find a Perimeter
Geometry Find the perimeter of
the square.
x6
3x
Solution
1. A square has four sides of equal length. Write an equation and
solve for x.
3x x 6
Write equation.
3x x x 6 x
2x 6
2x
Subtract x from each side.
Simplify.
6
2
Divide each side by 2 .
2
x 3
Simplify.
2. Find the length of one side by substituting 3 for x in either
expression.
( )
3x 3 3
9
Substitute for x and multiply.
3. To find the perimeter, multiply the length of one side by 4 .
4 p 9 36
Answer: The perimeter of the square is 36 units.
Checkpoint Find the perimeter of the square.
5.
3x 8
5x
80 units
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51
Focus On
Algebra
Rewriting Equations and Formulas
Use after Lesson 3.3 Goal: Rewrite literal equations and formulas.
Vocabulary
Literal
equation
Example 1
Solving a Literal Equation
Solve ax bx c for x. Then use the solution to solve 5x 2x 12.
Solution
1. Solve ax bx c for x.
Write original equation.
c
When solving
you calculate
When
a literala
slope,
be
sure
use
equation, followtothe
the x- steps
and y-you would
same
coordinates
of thea
take when solving
two points
in the of
specific
equation
same
order.
that form.
x
Distributive property
Assume
. Divide each side by
.
2. Use the solution to solve 5x 2x 12.
x
Solution of literal equation
Substitute
Simplify.
for a,
Answer: The solution of 5x 2x 12 is
for b, and
for c.
.
Checkpoint Solve the literal equation for x. Then use the
solution to solve the specific equation.
x
x
1. a b c; 1 3
4
52
Chapter 3 • Pre-Algebra Notetaking Guide
2. ax c bx; 5x 9 4x
Copyright © Holt McDougal. All rights reserved.
Focus On
Algebra
Rewriting Equations and Formulas
Use after Lesson 3.3 Goal: Rewrite literal equations and formulas.
Vocabulary
An equation such as ax b c, in which
coefficients and constants have been replaced by
letters
Literal
equation
Solving a Literal Equation
Example 1
Solve ax bx c for x. Then use the solution to solve 5x 2x 12.
Solution
1. Solve ax bx c for x.
When solving
you calculate
When
a literala
slope,
be
sure
use
equation, followtothe
the x- steps
and y-you would
same
coordinates
of thea
take when solving
two points
in the of
specific
equation
same
order.
that form.
ax bx c
Write original equation.
(a b)x c
Distributive property
x
c
Assume (a b) ≠ 0 . Divide each side by (a b) .
a b
2. Use the solution to solve 5x 2x 12.
x
c
Solution of literal equation
a b
12
Substitute 5 for a, 2 for b, and 12 for c.
5 2
4
Simplify.
Answer: The solution of 5x 2x 12 is 4 .
Checkpoint Solve the literal equation for x. Then use the
solution to solve the specific equation.
x
x
1. a b c; 1 3
4
x a(b c); 16
52
Chapter 3 • Pre-Algebra Notetaking Guide
2. ax c bx; 5x 9 4x
c
x ; 1
ab
Copyright © Holt McDougal. All rights reserved.
Rewriting an Equation
Example 2
Solve 3y 6x 12 for y.
Solution
3y 6x 12
Write original equation.
Add
to each side.
.
Multiply each side by
Use
and simplify.
Checkpoint Solve the equation for y.
3. 2y 8 14x
Example 3
4. 18 6x 9y
Rewriting and Using a Geometric Formula
The area A of a rectangle is given by the formula A lw where l is the
length and w is the width.
a. Solve the formula for the width w.
b. Use the rewritten formula to find the width of the rectangle shown.
A 640 ft2
w
I 40 ft
Solution
a. A lw
Original formula.
w
Notice in Example 1
that you isolate x by
A 640
ft2 w
working
backward.
First you subtract
I side
40 and
ft
from each
then you divide.
b. Substitute
each side by
for
w
and
for
.
in the rewritten formula.
Write rewritten formula.
Substitute
Simplify.
for
and
for
.
Answer: The width is
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Chapter 3 • Pre-Algebra Notetaking Guide
53
Example 2
Rewriting an Equation
Solve 3y 6x 12 for y.
Solution
3y 6x 12
Write original equation.
3y 6x 12
Add 6x to each side.
1
1
(3y) (6x 12)
3
3
Multiply each side by 1 .
y 2x 4
Use distributive property and simplify.
3
Checkpoint Solve the equation for y.
3. 2y 8 14x
4. 18 6x 9y
2
y 2 x
3
y 7x 4
Example 3
Rewriting and Using a Geometric Formula
The area A of a rectangle is given by the formula A lw where l is the
length and w is the width.
a. Solve the formula for the width w.
b. Use the rewritten formula to find the width of the rectangle shown.
A 640 ft2
w
I 40 ft
Solution
a. A lw
w
Notice in Example 1
that you isolate x by
A 640
ft2 w
working
backward.
First you subtract
I side
40 and
ft
from each
then you divide.
A
Original formula.
Divide each side by I .
I
b. Substitute 640 for A and 40 for I in the rewritten formula.
w
A
Write rewritten formula.
I
64 0
Substitute 640 for A and 40 for I .
16
Simplify.
40
Answer: The width is 16 feet.
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Chapter 3 • Pre-Algebra Notetaking Guide
53
3.4
Solving Inequalities Using
Addition or Subtraction
Goal: Solve inequalities using addition or subtraction.
Vocabulary
Inequality:
Solution of
an inequality:
Equivalent
inequalities:
Example 1
Writing and Graphing an Inequality
Air Travel An airline allows passengers to carry on-board one
piece of luggage. Luggage that exceeds 40 pounds cannot be
carried on-board. Write an inequality that gives the weight of
luggage that cannot be carried on-board.
Solution
Let w represent the weight of the luggage. Because the weight
cannot exceed 40 pounds, the weight of luggage that cannot be
carried on-board must be
.
Answer: The inequality is
0
54
. Draw the graph below.
10 20 30 40 50 60 70 80
Chapter 3 • Pre-Algebra Notetaking Guide
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3.4
Solving Inequalities Using
Addition or Subtraction
Goal: Solve inequalities using addition or subtraction.
Vocabulary
An inequality is a statement formed by placing an
Inequality: inequality symbol, such as < or >, between two
expressions.
The solution of an inequality with a variable is the
Solution of
set of all numbers that produce true statements
an inequality: when substituted for the variable.
Equivalent inequalities are inequalities that have
Equivalent
inequalities: the same solution.
Example 1
Writing and Graphing an Inequality
Air Travel An airline allows passengers to carry on-board one
piece of luggage. Luggage that exceeds 40 pounds cannot be
carried on-board. Write an inequality that gives the weight of
luggage that cannot be carried on-board.
Solution
Let w represent the weight of the luggage. Because the weight
cannot exceed 40 pounds, the weight of luggage that cannot be
carried on-board must be greater than 40 pounds .
Answer: The inequality is w > 40 . Draw the graph below.
0
54
10 20 30 40 50 60 70 80
Chapter 3 • Pre-Algebra Notetaking Guide
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Addition and Subtraction Properties of Inequality
Words Adding or subtracting the same number on each side of
an inequality produces an equivalent inequality.
The addition and
subtraction properties
of inequality are also
true for inequalities
involving ≤ and ≥.
Algebra If a < b, then a c < b c and a c < b c.
If a > b, then a c > b c and a c > b c.
Solving an Inequality Using Subtraction
Example 2
Solve m 9 ≤ 12. Graph and check your solution.
m 9 ≤ 12
m9
Write original inequality.
≤ 12 Subtract
m≤
Simplify.
Answer: The solution is m ≤
1
0
1
2
from each side.
3
4
5
6
. Draw the graph below.
7
Check: Choose any number less than or equal to
the number into the original inequality.
m 9 ≤ 12
Write original inequality.
?
9 ≤ 12
Substitute 0 for m.
12
.
Solving an Inequality Using Addition
Example 3
You can read an
inequality from left to
right as well as from
right to left. For
instance, "2 is
greater than x" can
also be read "x is
less than 2."
Algebraically, this
means that 2 x
can also be written
as x 2.
. Substitute
Solve 7 < x 11. Graph your solution.
7 < x 11
7 Write original inequality.
< x 11 Add
<x
Simplify.
< x, or
Answer: The solution is
Copyright © Holt McDougal. All rights reserved.
1
0
1
2
3
4
5
6
to each side.
. Draw the graph below.
7
Chapter 3 • Pre-Algebra Notetaking Guide
55
Addition and Subtraction Properties of Inequality
Words Adding or subtracting the same number on each side of
an inequality produces an equivalent inequality.
The addition and
subtraction properties
of inequality are also
true for inequalities
involving ≤ and ≥.
Algebra If a < b, then a c < b c and a c < b c.
If a > b, then a c > b c and a c > b c.
Solving an Inequality Using Subtraction
Example 2
Solve m 9 ≤ 12. Graph and check your solution.
m 9 ≤ 12
Write original inequality.
m 9 9 ≤ 12 9
Subtract 9 from each side.
m≤ 3
Simplify.
Answer: The solution is m ≤ 3 . Draw the graph below.
1
0
1
2
3
4
5
6
7
Check: Choose any number less than or equal to 3 . Substitute
the number into the original inequality.
m 9 ≤ 12
Write original inequality.
?
0 9 ≤ 12
≤ 12
9
Solution checks .
Solving an Inequality Using Addition
Example 3
You can read an
inequality from left to
right as well as from
right to left. For
instance, "2 is
greater than x" can
also be read "x is
less than 2."
Algebraically, this
means that 2 x
can also be written
as x 2.
Substitute 0 for m.
Solve 7 < x 11. Graph your solution.
7 < x 11
Write original inequality.
7 11 < x 11 11
Add 11 to each side.
4 <x
Simplify.
Answer: The solution is 4 < x, or x > 4 . Draw the graph below.
Copyright © Holt McDougal. All rights reserved.
1
0
1
2
3
4
5
6
7
Chapter 3 • Pre-Algebra Notetaking Guide
55
Checkpoint Solve the inequality. Graph and check your solution.
1. m 7 < 13
1
0
1
2
3
4
5
6
7
1
0
1
2
3
4
5
6
7
1
0
1
2
3
4
5
6
7
1
0
1
2
3
4
5
6
7
2. a 4 ≥ 5
3. x 2 ≤ 3
4. 6 < z 7
56
Chapter 3 • Pre-Algebra Notetaking Guide
Copyright © Holt McDougal. All rights reserved.
Checkpoint Solve the inequality. Graph and check your solution.
1. m 7 < 13
m<6
1
0
1
2
3
4
5
6
7
5
6
7
5
6
7
6
7
2. a 4 ≥ 5
a≥1
1
0
1
2
3
4
3. x 2 ≤ 3
x≤5
1
0
1
2
3
4
4. 6 < z 7
1 < z or z > 1
1
56
Chapter 3 • Pre-Algebra Notetaking Guide
0
1
2
3
4
5
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3.5
Solving Inequalities Using
Multiplication or Division
Goal: Solve inequalities using multiplication or division.
Multiplication Property of Inequality
Words Multiplying each side of an inequality by a positive
number produces an equivalent inequality.
Multiplying each side of an inequality by a negative number and
reversing the direction of the inequality symbol produces an
equivalent inequality.
The multiplication
properties of
inequality are also
true for inequalities
involving >, ≤, and ≥.
Algebra If a < b and c > 0, then ac
bc.
If a < b and c < 0, then ac
bc.
Solving an Inequality Using Multiplication
Example 1
m
4
Solve > 2.
m
> 2
4
Write original inequality.
m
4
Multiply each side by
.
Reverse inequality symbol.
p2
Simplify.
m
Checkpoint Solve the inequality. Graph your solution.
t
5
b
8
1. < 3
9
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10
2. ≤ 1
11
12
13
14
15
16
9 8 7 6 5 4 3 2
Chapter 3 • Pre-Algebra Notetaking Guide
57
3.5
Solving Inequalities Using
Multiplication or Division
Goal: Solve inequalities using multiplication or division.
Multiplication Property of Inequality
Words Multiplying each side of an inequality by a positive
number produces an equivalent inequality.
Multiplying each side of an inequality by a negative number and
reversing the direction of the inequality symbol produces an
equivalent inequality.
The multiplication
properties of
inequality are also
true for inequalities
involving >, ≤, and ≥.
Algebra If a < b and c > 0, then ac < bc.
If a < b and c < 0, then ac > bc.
Solving an Inequality Using Multiplication
Example 1
m
4
Solve > 2.
m
> 2
4
4
m
4
<
m <
Write original inequality.
4 p 2
Multiply each side by 4 .
Reverse inequality symbol.
8
Simplify.
Checkpoint Solve the inequality. Graph your solution.
t
5
b
8
1. < 3
2. ≤ 1
t < 15
9
Copyright © Holt McDougal. All rights reserved.
10
11
12
13
b ≥ 8
14
15
16
9 8 7 6 5 4 3 2
Chapter 3 • Pre-Algebra Notetaking Guide
57
Division Property of Inequality
Words Dividing each side of an inequality by a positive number
produces an equivalent inequality.
Dividing each side of an inequality by a negative number and
reversing the direction of the inequality symbol produces an
equivalent inequality.
The division
properties of
inequality are also
true for inequalities
involving >, ≤, and ≥.
a
c
a
If a < b and c < 0, then c
Algebra If a < b and c > 0, then b
.
c
b
.
c
Solving an Inequality Using Division
Example 2
Solve 11t ≥ 33.
11t ≥ 33
11t
Write original inequality.
33
Divide each side by
.
Reverse inequality symbol.
Simplify.
t
Checkpoint Solve the inequality. Graph your solution.
3. 4y ≤ 36
3
58
4
5
Chapter 3 • Pre-Algebra Notetaking Guide
4. 3x > 12
6
7
8
9
10
9 8 7 6 5 4 3 2
Copyright © Holt McDougal. All rights reserved.
Division Property of Inequality
Words Dividing each side of an inequality by a positive number
produces an equivalent inequality.
Dividing each side of an inequality by a negative number and
reversing the direction of the inequality symbol produces an
equivalent inequality.
The division
properties of
inequality are also
true for inequalities
involving >, ≤, and ≥.
a
b
c
c
a
b
If a < b and c < 0, then > .
c
c
Algebra If a < b and c > 0, then < .
Solving an Inequality Using Division
Example 2
Solve 11t ≥ 33.
11t ≥ 33
11t
11
Write original inequality.
33
≤ 11
t ≤
3
Divide each side by 11 .
Reverse inequality symbol.
Simplify.
Checkpoint Solve the inequality. Graph your solution.
3. 4y ≤ 36
4. 3x > 12
y≤9
3
58
4
5
Chapter 3 • Pre-Algebra Notetaking Guide
6
7
x < 4
8
9
10
9 8 7 6 5 4 3 2
Copyright © Holt McDougal. All rights reserved.
3.6
Solving Multi-Step Inequalities
Goal: Solve multi-step inequalities.
Example 1
Writing and Solving a Multi-Step Inequality
Charity Walk You are participating in a charity walk. You want
to raise at least $500 for the charity. You already have $175 by
asking people to pledge $25 each. How many more $25 pledges
do you need?
Solution
Let p represent the number of additional pledges. Write a verbal
model.
p
≥
≥
≥
≥ p≥
Answer: You need at least
≥
Substitute.
Subtract
each side.
from
Simplify.
Divide each side
by
.
Simplify.
more $25 pledges.
Checkpoint
1. Look back at Example 1. Suppose you wanted to raise at least
$620 and you already have raised $380 by asking people to
pledge $20 each. How many more $20 pledges do you need?
Copyright © Holt McDougal. All rights reserved.
Chapter 3 • Pre-Algebra Notetaking Guide
59
3.6
Solving Multi-Step Inequalities
Goal: Solve multi-step inequalities.
Example 1
Writing and Solving a Multi-Step Inequality
Charity Walk You are participating in a charity walk. You want
to raise at least $500 for the charity. You already have $175 by
asking people to pledge $25 each. How many more $25 pledges
do you need?
Solution
Let p represent the number of additional pledges. Write a verbal
model.
Additional
Amount
Minimum
Money already
p
≥
pledges
per pledge
desired amount
raised
175 25p ≥ 500
Substitute.
175 25p 175 ≥ 500 175
25p ≥ 325
25p
325
≥ 25
25
p ≥ 13
Subtract 175 from
each side.
Simplify.
Divide each side
by 25 .
Simplify.
Answer: You need at least 13 more $25 pledges.
Checkpoint
1. Look back at Example 1. Suppose you wanted to raise at least
$620 and you already have raised $380 by asking people to
pledge $20 each. How many more $20 pledges do you need?
at least 12 more $20 pledges
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Chapter 3 • Pre-Algebra Notetaking Guide
59
Solving a Multi-Step Inequality
Example 2
x
9 < 7
3
x
9 3
Original inequality
< 7 Add
x
<
3
to each side.
Simplify.
x3 p
Multiply each side by
.
Reverse inequality symbol.
Simplify.
x
Checkpoint Solve the inequality. Then graph the solution.
x
4
2. 2x 9 < 25
3
4
5
6
3. 3 ≥ 2
7
8
9
10
4. 2 > 4 x
9 8 7 6 5 4 3 2
60
Chapter 3 • Pre-Algebra Notetaking Guide
3
4
5
6
7
8
9
10
6
7
8
9
10
x
2
5. 4 ≤ 9
3
4
5
Copyright © Holt McDougal. All rights reserved.
Solving a Multi-Step Inequality
Example 2
x
9 < 7
3
Original inequality
x
9 9 < 7 9
3
x
< 2
3
3
x3 >
x >
Add 9 to each side.
Simplify.
3 p 2
Multiply each side by 3 .
Reverse inequality symbol.
6
Simplify.
Checkpoint Solve the inequality. Then graph the solution.
x
4
2. 2x 9 < 25
3. 3 ≥ 2
x<8
3
4
5
6
7
x≥4
8
9
10
4. 2 > 4 x
3
4
5
60
Chapter 3 • Pre-Algebra Notetaking Guide
7
8
9
10
8
9
10
x
2
5. 4 ≤ 9
x > 6
9 8 7 6 5 4 3 2
6
x ≤ 10
3
4
5
6
7
Copyright © Holt McDougal. All rights reserved.
3
Words to Review
Give an example of the vocabulary word.
Inequality
Solution of an inequality
Equivalent inequalities
Literal equation
Review your notes and Chapter 3 by using the Chapter Review on
pages 156–159 of your textbook.
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Chapter 3 • Pre-Algebra Notetaking Guide
61
3
Words to Review
Give an example of the vocabulary word.
Inequality
x>0
Equivalent inequalities
2x < 10 and x > 5
Solution of an inequality
The solution of m 9 ≤ 12 is
m ≤ 3.
Literal equation
ax by c
Review your notes and Chapter 3 by using the Chapter Review on
pages 156–159 of your textbook.
Copyright © Holt McDougal. All rights reserved.
Chapter 3 • Pre-Algebra Notetaking Guide
61