CONFIDENTIAL 1

Grade 8 Pre-AlgebraGrade 8 Pre-Algebra

Solving Equations withSolving Equations withVariables on Both SidesVariables on Both Sides

CONFIDENTIAL 2

Warm UpWarm Up

1) 15y + 31 = 61

2) 9 - c = -13

3) x + 4= 15 6

4) y + 1 = 5 3 4 12

Solve each equation. Check your answer:

1) y = 2

2) c = 22

3) x = 66

4) y = 0.5

CONFIDENTIAL 3

Model Equations with Variables on Both Sides

You can use algebra tiles to understand how to solve equations with variables on both sides.

KEY REMEMBER

= 1

= -1= x + = 0= -x

CONFIDENTIAL 4

Activity

Use algebra tiles to model and solve 5x - 2 = 2x+ 10.

MODEL ALGEBRA

5x - 2 = 2x + 10

5x - 2 - 2x = 2x - 2x + 10

3x - 2 = 10

Next page

Model 5x - 2 on the left side of the mat and 2x + 10 on the right side. Remember that 5x - 2 is the same as 5x + (-2) .

Remove 2 x-tiles from both sides. This represents subtracting 2x from both sides of the equation.

CONFIDENTIAL 5

MODEL ALGEBRA

3x - 2 + 2 = 10 + 2

3x = 12

3x = 12 3 3

x = 4

Use algebra tiles to model and solve the equation.1. 5x + 12 = 2x + 3

Remove Place 2 yellow tiles on both sides. This represents adding 2 to both sides of the equation. Remove zero pairs.

Separate each side into 3 equal groups. Each group is 1/3 of the side. One x-tile is equivalent to 4 yellow tiles.

CONFIDENTIAL 6

To solve an equation with variables on both sides, useinverse operations to “collect” variable

terms on one side of the equation.

Solving Equations with Variables on Both Sides

Solve each equation.

a) 7k = 4k + 15

7k = 4k + 15-4k -4k

3k = 15To collect the variable terms on one side, subtract 4k from both sides.

Since k is multiplied by 3, divide both sides by 3 to undo the multiplication.

3k = 15 3 3

k = 5

CONFIDENTIAL 7

b) 5x - 2 = 3x + 4

5x - 2 = 3x + 4- 3x -3x

2x - 2 = 4

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

Check

To check your solution, substitute 3 for x in the original equation.

+ 2 +22x = 6

Since 2 is subtracted from 2x, add 2 to both sides to undo the subtraction.

2x = 6 2 2

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

x = 3

5x - 2 = 3x + 45 (3) - 2 15 - 2 9 + 4 13 13

3 (3) + 4

CONFIDENTIAL 8

Now you try!

1) 4b + 2 = 3b

2) 0.5 + 0.3y = 0.7y - 0.3

Solve each equation. Check your answer.

To solve more complicated equations, you may need to first simplify by using the Distributive

Property or combining like terms.

1) b = -2

2) y = 2

CONFIDENTIAL 9

Simplifying Each Side Before Solving EquationsSolve each equation.

A) 2(y + 6) = 3y

Distribute 2 to the expression in parentheses.

2y + 12 = 3y- 2y -2y

12 = y

Check

To check your solution, substitute 12 for y in the original equation.

2(y + 6) = 3y2(y) + 2(6) = 3y

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

2(y + 6) = 3y2(12 + 6) 3(12)

2(18) 3636 36

CONFIDENTIAL 10

B) 3 - 5b + 2b = -2 - 2(1 - b)

Distribute -2 to the expression in parentheses.

3 = -4 + 5b+3b +3b

Since b is multiplied by 5, divide both sides by 5.

3 - 5b + 2b = -2 - 2(1 - b)3 - 5b + 2b = -2 - 2(1) - 2(-b)3 - 5b + 2b = -2 - 2 + 2b

3 - 3b = -4 + 2b Combine like terms.

Add 3b to both sides.

Since -4 is added to 5b, add 4 to both sides.+4 +4

7 = 5b7 = 5b5 5

1.4 = b

CONFIDENTIAL 11

Now you try!

1) 1 (b + 6) = 3b – 1 2 2

2) 3x + 15 - 9 = 2 (x + 2)

Solve each equation. Check your answer.

1) b = 4

2) x = -2

CONFIDENTIAL 12

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.

Identities and Contradictions

WORDS NUMBERS ALGEBRAIdentity: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.

2 + 1 = 2 + 1

3 = 3

2 + x = 2 + x- x - x 2 = 2

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

1 = 1 + 2

1 = 3

x = x + 3- x - x0 = 3

CONFIDENTIAL 13

Infinitely Many Solutions or No Solutions

A) x + 4 - 6x = 6 - 5x - 2

Identify like terms.

+ 5x + 5x4 = 4

Solve each equation.

x + 4 - 6x = 6 - 5x - 2

4 - 5x = 4 - 5x Combine like terms on the left and the right.Add 5x to both sides.True statement

The equation x + 4 - 6x = 6 - 5x - 2 is an identity. All values of x will make the equation true. All real

numbers are solutions.

CONFIDENTIAL 14

A) -8x + 6 + 9x = -17 + x

Identify like terms.

- x - x6 = -17

-8x + 6 + 9x = -17 + x

6 + x = -17 + x Combine like terms .

Subtract x to both sides.

False statement

The equation -8x + 6 + 9x = -17 + x is a contradiction. There is no value of x that will

make the equation true. There are no solutions.

CONFIDENTIAL 15

Now you try!

1) 4y + 7 - y = 10 + 3y

2) 2c + 7 + c = -14 + 3c + 21

Check each equation.

1) False statement

2) True statement

CONFIDENTIAL 16

When is 36¢ plus 3¢ per minute the same as 6¢ per minute ?

Solution:

36 + 3m = 6m To collect the variable terms on one side, subtract 3m from both sides.

Phone Company ChargesCompany A 36¢ plus 3¢ per minuteCompany B 6¢ per minute

Let m represent minutes, and write expressions for each company’s cost.

The long-distance rates of two phone companies are shown in the table. How long is a call that costs the same amount no matter which company is used?

What is the cost of that call?

- 3m - 3m36 = 3m

Next page

CONFIDENTIAL 17

12 = m

Since m is multiplied by 3, divide both sides by 3 to undo the multiplication.

The charges will be the same for a 12-minute call using either phone service. To find the cost of this call, evaluate either expression for m = 12:

The cost of a 12-minute call through either company is 72¢.

36 = 3m 3 3

36 + 3m = 36 + 3 (12) = 36 + 36 = 72

6m = 6 (12) = 72

CONFIDENTIAL 18

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

Now you try!

10 years.

CONFIDENTIAL 19

Assessment

1) 2c - 5 = c + 4

2) 8r + 4 = 10 + 2r

3) 2x -1 = x + 11

4) 28 - 0.3y = 0.7y - 12

Factor each trinomial by guess and check:

1) c = 9

2) r = 13) x = 12

4) y = 40

CONFIDENTIAL 20

5) -2 (x + 3) = 4x -3

6) 3c - 4c + 1 = 5c + 2 + 3

7) 5 + 3(q - 4) = 2(q + 1)

8) 5 - (t + 3) = -1 + 2(t -3)

9) 7x - 4 = -2x + 1 + 9x - 5

Factor the trinomial. Check your answer.

5) x = -0.5

6) c = -0.67

7) q = 9

8) t = 3

9) True statement

CONFIDENTIAL 21

10) A house-painting company charges $376 plus $12 per hour. Another painting company charges

$280 plus $15 per hour.

a. How long is a job for which both companies will charge the same amount?

b. What will that cost be?

10) a. 32 hours b. $760

CONFIDENTIAL 22

To solve an equation with variables on both sides, useinverse operations to “collect” variable

terms on one side of the equation.

Solving Equations with Variables on Both Sides

Solve each equation.a) 7k = 4k + 15

7k = 4k + 15-4k -4k

3k = 15To collect the variable terms on one side, subtract 4k from both sides.

Since k is multiplied by 3, divide both sides by 3 to undo the multiplication.

3k = 15 3 3k = 5

Let’s review

CONFIDENTIAL 23

b) 5x - 2 = 3x + 4

5x - 2 = 3x + 4- 3x -3x

2x - 2 = 4

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

Check

To check your solution, substitute 3 for x in the original equation.

+ 2 +22x = 6

Since 2 is subtracted from 2x, add 2 to both sides to undo the subtraction.

2x = 6 2 2

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

x = 3

5x - 2 = 3x + 45 (3) - 2 15 - 2 9 + 4 13 13

3 (3) + 4

review

CONFIDENTIAL 24

Simplifying Each Side Before Solving Equations

A) 2(y + 6) = 3y

Distribute 2 to the expression in parentheses.

2y + 12 = 3y- 2y -2y

12 = yCheck

To check your solution, substitute 12 for y in the original equation.

2(y + 6) = 3y2(y) + 2(6) = 3y

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

2(y + 6) = 3y2(12 + 6) 3(12)

2(18) 3636 36

review

CONFIDENTIAL 25

B) 3 - 5b + 2b = -2 - 2(1 - b)

Distribute -2 to the expression in parentheses.

3 = -4 + 5b+3b +3b

Since b is multiplied by 5, divide both sides by 5.

3 - 5b + 2b = -2 - 2(1 - b)3 - 5b + 2b = -2 - 2(1) - 2(-b)3 - 5b + 2b = -2 - 2 + 2b

3 - 3b = -4 + 2b Combine like terms.

Add 3b to both sides.

Since -4 is added to 5b, add 4 to both sides.+4 +4

7 = 5b7 = 5b5 5

1.4 = b

review

CONFIDENTIAL 26

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.

Identities and Contradictions

WORDS NUMBERS ALGEBRAIdentity: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.

2 + 1 = 2 + 1

3 = 3

2 + x = 2 + x- x - x 2 = 2

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

1 = 1 + 2

1 = 3

x = x + 3- x - x0 = 3

review

CONFIDENTIAL 27

Infinitely Many Solutions or No Solutions

A) x + 4 - 6x = 6 - 5x - 2

Identify like terms.

+ 5x + 5x4 = 4

Solve each equation.

x + 4 - 6x = 6 - 5x - 2

4 - 5x = 4 - 5x Combine like terms on the left and the right.Add 5x to both sides.True statement

The equation x + 4 - 6x = 6 - 5x - 2 is an identity. All values of x will make the equation true. All real

numbers are solutions.

review

CONFIDENTIAL 28

A) -8x + 6 + 9x = -17 + x

Identify like terms.

- x - x6 = -17

-8x + 6 + 9x = -17 + x

6 + x = -17 + x Combine like terms .

Subtract x to both sides.

False statement

The equation -8x + 6 + 9x = -17 + x is a contradiction. There is no value of x that will

make the equation true. There are no solutions.

review

CONFIDENTIAL 29

When is 36¢ plus 3¢ per minute the same as 6¢ per minute ?

Solution:

36 + 3m = 6m To collect the variable terms on one side, subtract 3m from both sides.

Phone Company ChargesCompany A 36¢ plus 3¢ per minuteCompany B 6¢ per minute

Let m represent minutes, and write expressions for each company’s cost.

The long-distance rates of two phone companies are shown in the table. How long is a call that costs the same amount no matter which company is used?

What is the cost of that call?

- 3m - 3m36 = 3m

Next page

review

CONFIDENTIAL 30

12 = m

Since m is multiplied by 3, divide both sides by 3 to undo the multiplication.

The charges will be the same for a 12-minute call using either phone service. To find the cost of this call, evaluate either expression for m = 12:

The cost of a 12-minute call through either company is 72¢.

36 = 3m 3 3

36 + 3m = 36 + 3 (12) = 36 + 36 = 72

6m = 6 (12) = 72

review

CONFIDENTIAL 31

You did a great job You did a great job today!today!