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Holt McDougal Algebra 1

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Solving Equations with

Variables on Both Sides

Solving Equations with

Holt Algebra 1

Warm Up

Lesson Presentation

Lesson Quiz

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Warm Up

Simplify. 1. 4x – 10x

2. –7(x – 3)

3.

4. 15 – (x – 2)

Solve.

5. 3x + 2 = 8

6.

–6x

–7x + 21

17 – x

2x + 3

2

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Holt McDougal Algebra 1

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Solving Equations with

Solve equations in one variable that

contain variable terms on both sides.

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Vocabulary

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Holt McDougal Algebra 1

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Solving Equations with

To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

Helpful Hint

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Solving Equations with

Solve 7n – 2 = 5n + 6.

Example 1: Solving Equations with Variables on Both Sides

To collect the variable terms on one side, subtract 5n from both sides.

7n – 2 = 5n + 6 –5n –5n

2n – 2 = 6

Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.

2n = 8

+ 2 + 2

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Solving Equations with

Solve 4b + 2 = 3b.

Check It Out! Example 1a

To collect the variable terms on one side, subtract 3b from both sides.

4b + 2 = 3b

–3b –3b

b + 2 = 0

b = –2

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Solving Equations with

Solve 0.5 + 0.3y = 0.7y – 0.3.

Check It Out! Example 1b

To collect the variable terms on one side, subtract 0.3y from both sides.

0.5 + 0.3y = 0.7y – 0.3 –0.3y –0.3y

0.5 = 0.4y – 0.3

0.8 = 0.4y

+0.3 + 0.3

2 = y

Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction.

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Solving Equations with

To solve more complicated

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Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2: Simplifying Each Side Before Solving Equations

Combine like terms. Distribute –5 to the

expression in parentheses.

4 – 6a + 4a = –1 –5(7 – 2a)

4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a

4 – 2a = –36 + 10a

+36 +36

40 – 2a = 10a

+ 2a +2a 40 = 12a

Since –36 is added to 10a, add 36 to both sides.

To collect the variable

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Solving Equations with

Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2 Continued

40 = 12a

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Solve .

Since 1 is subtracted from b, add 1 to both sides.

Distribute to the expression in parentheses.

1 2

+ 1 + 1

3 = b – 1

To collect the variable terms on one side, subtract b from both sides.

1 2

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Solving Equations with

Solve 3x + 15 – 9 = 2(x + 2).

Combine like terms.

Distribute 2 to the expression in parentheses.

3x + 15 – 9 = 2(x + 2)

3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4

3x + 6 = 2x + 4

–2x –2x

x + 6 = 4

– 6 – 6

x = –2

To collect the variable terms on one side, subtract 2x from both sides.

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Solving Equations with

An

identity

is an equation that is true

for all values of the variable. An equation

that is an identity has infinitely many

solutions.

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Holt McDougal Algebra 1

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Solving Equations with

WORDS

Identity

When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has

infinitely many solutions.

NUMBERS

2 + 1 = 2 + 1

3 = 3



ALGEBRA

2 + x = 2 + x

–

x

–

x

2 = 2



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False Equations

When solving an equation, if you get a false equation, the original equation has no solutions.

WORDS

x = x + 3

–

x

–

x

0 = 3



1 = 1 + 2

1 = 3



ALGEBRA NUMBERS

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Holt McDougal Algebra 1

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Solving Equations with

Solve 10 – 5x + 1 = 7x + 11 – 12x.

Example 3A: Infinitely Many Solutions or No Solutions

Add 5x to both sides.

Identify like terms.

10 – 5x + 1 = 7x + 11 – 12x

11 – 5x = 11 – 5x

11 = 11

+ 5x + 5x

True statement.

Combine like terms on the left and the right.

10 – 5x + 1 = 7x + 11 – 12x



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Solving Equations with

Solve 12x – 3 + x = 5x – 4 + 8x.

Example 3B: Infinitely Many Solutions or No Solutions

Subtract 13x from both sides. Identify like terms.

12x – 3 + x = 5x – 4 + 8x

13x – 3 = 13x – 4

–3 = –4

–13x –13x

False statement.

12x – 3 + x = 5x – 4 + 8x



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Solving Equations with

Solve 4y + 7 – y = 10 + 3y.

Subtract 3y from both sides. Identify like terms.

4y + 7 – y = 10 + 3y

3y + 7 = 3y + 10

7 = 10

–3y –3y

False statement.

4y + 7 – y = 10 + 3y



The equation 4y + 7 – y = 10 + 3y is a false

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Solving Equations with

Solve 2c + 7 + c = –14 + 3c + 21.

Subtract 3c both sides. Identify like terms.

2c + 7 + c = –14 + 3c + 21

3c + 7 = 3c + 7

7 = 7

–3c –3c

True statement.

2c + 7 + c = –14 + 3c + 21



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Solving Equations with

Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In

how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs

will that be?

Example 4: Application

Person Bulbs

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Solving Equations with

Example 4: Application Continued

Let b represent bulbs, and write expressions for the number of bulbs planted.

60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour

When is ?

60 + 44b = 96 + 32b

– 32b – 32b

To collect the variable terms on one side, subtract 32b from both sides.

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Holt McDougal Algebra 1

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Solving Equations with

Since 60 is added to 12b, subtract 60 from both sides.

60 + 12b = 96

–60 – 60

12b = 36

Since b is multiplied by 12, divide both sides by 12 to undo the multiplication.

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After 3 hours, Jon and Sara will have planted the

same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either

expression for b = 3:

60 + 44b = 60 + 44(3) = 60 + 132 = 192 96 + 32b = 96 + 32(3) = 96 + 96 = 192

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Holt McDougal Algebra 1

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Solving Equations with

Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg?

Check It Out! Example 4

Let g represent Greg's age, and write expressions for his age.

four times Greg's

age

decreased

by 3

is equal to three times Greg's age increased

by 7 .

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Solving Equations with

Check It Out! Example 4 Continued

4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides.

g – 3 = 7

–3g –3g

Since 3 is subtracted from g, add 3 to both sides.

+ 3 + 3

g = 10

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Holt McDougal Algebra 1

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Solving Equations with

Lesson Quiz

Solve each equation.

1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4

3. 6 – 7(a + 1) = –3(2 – a)

4. 4(3x + 1) – 7x = 6 + 5x – 2

5.

6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same?

3 8

all real numbers

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