UNIT 2.3 MULTI STEP EQUATIONSUNIT 2.3 MULTI STEP EQUATIONS

Warm UpSimplify. 1. 4x – 10x2. –7(x – 3)

3.

4. 15 – (x – 2)Solve.5. 3x + 2 = 8

6.

–6x–7x + 21

17 – x

2x + 3

2

28

Solve equations in one variable that contain variable terms on both sides.

Objective

Vocabularyidentitycontradiction

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

Helpful HintEquations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.

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

n = 4

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 – 2 – 2

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.

Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.

To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2: Simplifying Each Side BeforeSolving 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 terms on one side, add 2a to both sides.

Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2 Continued

40 = 12aSince a is multiplied by 12,

divide both sides by 12.

Solve .

Check It Out! Example 2A

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

Distribute to the expression in parentheses.

12

+ 1 + 13 = b – 1

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

12

4 = b

Solve 3x + 15 – 9 = 2(x + 2).

Check It Out! Example 2B

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.

Since 6 is added to x, subtract 6 from both sides to undo the addition.

An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions.

A contradiction is an equation that is not true for any value of the variable. It has no solutions.

WORDS

IdentityWhen solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions.

NUMBERS2 + 1 = 2 + 1

3 = 3

ALGEBRA

2 + x = 2 + x –x –x 2 = 2

Identities and Contradictions

ContradictionWhen solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions.

WORDS

x = x + 3 –x –x 0 = 3

1 = 1 + 2 1 = 3

ALGEBRA

NUMBERS

Identities and Contradictions

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

The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

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.

Combine like terms on the left and the right.

12x – 3 + x = 5x – 4 + 8x

The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

Solve 4y + 7 – y = 10 + 3y.

Check It Out! Example 3a

Subtract 3y from both sides.

Identify like terms.4y + 7 – y = 10 + 3y

3y + 7 = 3y + 10

7 = 10 –3y –3y

False statement.

Combine like terms on the left and the right.

4y + 7 – y = 10 + 3y

The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.

Solve 2c + 7 + c = –14 + 3c + 21.

Check It Out! Example 3b

Subtract 3c both sides.

Identify like terms.2c + 7 + c = –14 + 3c + 21

3c + 7 = 3c + 7

7 = 7 –3c –3c

True statement.

Combine like terms on the left and the right.

2c + 7 + c = –14 + 3c + 21

The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

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 BulbsJon 60 bulbs plus 44 bulbs per hour

Sara 96 bulbs plus 32 bulbs per hour

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

60 + 44b = 96 + 32b– 32b – 32b

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

60 + 12b = 96

Example 4: Application Continued

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

60 + 12b = 96–60 – 60

12b = 36Since b is multiplied by 12,

divide both sides by 12 to undo the multiplication.

b = 3

Example 4: Application Continued

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

After 3 hours, Jon and Sara will each have planted 192 bulbs.

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

3is

equal to

three times Greg's age

increased by

7 .

4g – 3 = 3g + 7

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

Greg is 10 years old.

Lesson QuizSolve 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

1

20 hours

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