Pre-Algebra

10-3 Solving Equations with Variables on Both Sides10-3 Solving Equations with

Variables on Both Sides

Pre-Algebra

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Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

• What are some observations you made about solving equations with variables on both sides, based on last night’s homework?

• Goal is to isolate the variable – get variables on one side and numbers on the other.

• You can go in different order. There are different ways to go about the problem.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Ex A)

4x + 6 = x4x + 6 = x

– 4x – 4x

6 = –3x

–2 = x

6–3

–3x–3=

–x -6 – x -6

3x = –6

x = -2

3

3

4x + 6 = x

What can NOT do?

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Ex B)

9b – 6 = 5b + 189b – 6 = 5b + 18

– 5b – 5b

4b – 6 = 18

4b 4

24 4 =

b = 6

+ 6 + 6

4b = 24

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Ex.C)

9w + 3 = 5w + 7 + 4w9w + 3 = 5w + 7 + 4w

3 ≠ 7

9w + 3 = 9w + 7

– 9w – 9w

No solution. There is no number that can be substituted for the variable w to make the equation true.

With each step, you are keeping the equation, ajd just simplifying. If your final answer turns out to be not true, there is NO

SOLUTION that would work for this equation.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Ex. D

Multiply by the LCD.

4y + 12y – 15 = 20y – 14

16y – 15 = 20y – 14 Combine like terms.

y5

34

3y5

710

+ – = y –

20( ) = 20( )y5

34

3y5

710

+ – y –

20( ) + 20( ) – 20( )= 20(y) – 20( )y5

3y5

34

710

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Add 14 to both sides.

–15 = 4y – 14

–1 = 4y

+ 14 + 14

–1 4

4y4 = Divide both sides by 4.

-14 = y

16y – 15 = 20y – 14

– 16y – 16y Subtract 16y from both sides.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Ex.E)

12z – 12 – 4z = 6 – 2z + 3212z – 12 – 4z = 6 – 2z + 32

+ 12 +12

8z – 12 = –2z + 38+ 2z + 2z

10z – 12 = + 38

10z = 50

z = 5

10z 5010 10=

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3: Consumer Application

Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3 Continued

First solve for the price of one doughnut.

1.25 + 2d = 0.50 + 5dd = price of one doughnut.

– 2d – 2d

1.25 = 0.50 + 3d

– 0.50 – 0.50

0.75 = 3d

0.753

3d3=

0.25 = d The price of one doughnut is $0.25.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3 Continued

Now find the amount of money Jamie spends each morning.

1.25 + 2d Choose one of the original expressions.

Jamie spends $1.75 each morning.

1.25 + 2(0.25) = 1.75

0.25n0.25

1.75 0.25 =

Let n represent the number of doughnuts.

Find the number of doughnuts Jamie buys on Tuesday.

0.25n = 1.75

n = 7; Jamie bought 7 doughnuts on Tuesday.

Divide both sides by 0.25.