4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12.

5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.

California Standards

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.

Solving Equations with Variables on Both Sides

1: Move variable to one side7n – 2 = 5n + 6

–5n –5n

2n – 2 = 6

3: multiple/divide2n = 8

+ 2 + 2

n = 4

2: Add/subtract

Solve the equation. Check your answer.

1: Move variable to one side4b + 2 = 3b

–3b –3b

b + 2 = 0

b = –2

– 2 – 2

4b + 2 = 3b

2: add/subtract

4b + 2 = 3b Check

4(–2) + 2 3(–2)

–8 + 2 –6

–6 –6

0.5 + 0.3y = 0.7y – 0.3

1: Move variable to one side0.5 + 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

2: add/subtract

3: Multiple/divide

Solve the equation. Check your answer.

0.5 + 0.3y = 0.7y – 0.3 Check

0.5 + 0.3(2) 0.7(2) – 0.3

1.1 1.1

0.5 + 0.6 1.4 – 0.3

1.4a – 3 = 2a + 7

2.-3r + 9 = -4r + 5

3.3d + 8 = 2d – 17

4.2x – 7 = 5x – 10

a = 5

r = -4

d = -25

x = 1

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

Don’t call me after midnight

1. D= Distributive property

2. C= combine like term

3. M = move variable to one side

4. A = addition/subtraction

5. M = Multiplication/division

3x + 15 – 9 = 2(x + 2)

C: Combine like term

D: distributive Property

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

M: move variable to one

side

A: add/subtract

Solve the equation. Check your answer.

Simplifying Each Side BeforeSolving Equations

C: Combine like terms

D: Distributive P.

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

+2a +2a

4 = -36 + 12a

M: move variable to

One side

4 – 6a + 4a = –1 – 5(7 – 2a)

M: multiply/divide

4 = -36 + 12a+ 36 +36

40 = 12a

A: add/subtract

Check

3x + 15 – 9 = 2(x + 2)

3(–2) + 15 – 9 2(–2 + 2)

–6 + 15 – 9 2(0)

0 0

An identity is an equation that is always true, no matter what value is substituted for the variable. The solution set of an identity is all real numbers. Some equations are always false. Their solution sets are empty. In other words, their solution sets contain no elements.

Solve the equation.

Infinitely Many Solutions or No Solutions

11 – 5x = 11 – 5x

C: combine like

term

The statement 11 – 5x = 11 – 5x is true for all values of x. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. In other words, all real numbers are solutions.

10 – 5x + 1 = 7x + 11 – 12x

10 – 5x + 1 = 7x + 11 – 12x

Infinitely Many Solutions or No Solutions

M: move variable to

one side

13x – 3 = 13x – 4

–3 = –4

–13x –13x

C: combine like term

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

Solve the equation.

12x – 3 + x = 5x – 4 + 8x

12x – 3 + x = 5x – 4 + 8x

The empty set can be written as or {}.

Writing Math

Solve the equation.

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; the

solution set is .

4y + 7 – y = 10 + 3y

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

Solve the equation.

Check It Out! Example 3b

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

3c + 7 = 3c + 7 Combine like terms on the left and the right.

2c + 7 + c = –14 + 3c + 21

The statement 3c + 7 = 3c + 7 is true for all values of c. The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. In other words, 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?

Additional Example 4: Application

Person Bulbs

Jon 60 bulbs plus 44 bulbs per hour

Sara 96 bulbs plus 32 bulbs per hour

Additional Example 4 Continued

Let h represent hours, 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 + 44h = 96 + 32h

60 + 44h = 96 + 32h To collect the variable terms on one side, subtract 32h from both sides.

60 + 12h = 96

–32h –32h

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

60 + 12h = 96–60 – 60

12h = 36Since h is multiplied by 12,

divide both sides by 12 to undo the multiplication.

h = 3

Additional Example 4 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 h = 3:

60 + 44h = 60 + 44(3) = 60 + 132 = 192

96 + 32h = 96 + 32(3) = 96 + 96 = 192

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

Additional Example 4 Continued

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

120 hours