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Learn to solve equations with variables on both sides of the equal sign.PowerPoint Presentation

Learn to solve equations with variables on both sides of the equal sign.

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Learn to solve equations with variables on both sides of the equal sign.

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Learn to solve equations with variables on both sides of the equal sign.

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Learn to solve equations with variables on both sides of the equal sign.

Some problems produce equations that have variables on both sides of the equal sign.

Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

–3x

=

–3

6

–3

Additional Example 1A: Solving Equations with Variables on Both Sides

Solve.

4x + 6 = x

4x + 6 = x

– 4x– 4x

Subtract 4x from both sides.

6 = –3x

Divide both sides by –3.

–2 = x

Helpful Hint

Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2.

4b

24

=

4

4

Additional Example 1B: Solving Equations with Variables on Both Sides

Solve.

9b – 6 = 5b + 18

9b – 6 = 5b + 18

– 5b– 5b

Subtract 5b from both sides.

4b – 6 = 18

+ 6+ 6

Add 6 to both sides.

4b = 24

Divide both sides by 4.

b = 6

9w + 3 = 9w + 7

– 9w– 9w

Subtract 9w from both sides.

Additional Example 1C: Solving Equations with Variables on Both Sides

Solve.

9w + 3 = 9w + 7

3 ≠ 7

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

Helpful Hint

If the variables in an equation are eliminated and the resulting statement is false, the equation has no solution.

–4x

=

–4

8

–4

Check It Out: Example 1A

Solve.

5x + 8 = x

5x + 8 = x

– 5x– 5x

Subtract 5x from both sides.

8 = –4x

Divide both sides by –4.

–2 = x

Check It Out: Example 1B

Solve.

3b – 2 = 2b + 12

3b – 2 = 2b + 12

– 2b– 2b

Subtract 2b from both sides.

b – 2 = 12

+ 2+ 2

Add 2 to both sides.

b = 14

3w + 1 = 3w + 8

– 3w– 3w

Subtract 3w from both sides.

Check It Out: Example 1C

Solve.

3w + 1 = 3w + 8

1 ≠ 8

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

To solve multi-step equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

8z8

=

8

8

Additional Example 2: Solving Multi-Step Equations with Variables on Both Sides

Solve.

10z – 15 – 4z = 8 – 2z - 15

10z – 15 – 4z = 8 – 2z – 15

6z– 15 = –2z– 7

Combine like terms.

+ 2z+ 2z

Add 2z to both sides.

8z – 15 = – 7

+ 15+15

Add 15 to both sides.

8z = 8

Divide both sides by 8.

z = 1

10z50

=

10

10

Check It Out: Example 2

Solve.

12z – 12 – 4z = 6 – 2z + 32

12z – 12 – 4z = 6 – 2z + 32

8z– 12 = –2z+ 38

Combine like terms.

+ 2z+ 2z

Add 2z to both sides.

10z – 12 = 38

+ 12+12

Add 12 to both sides.

10z = 50

Divide both sides by 10.

z = 5

Additional Example 3: Business Application

Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florists’ bouquets cost the same price.

13.95

1.55r1.55

=

1.55

Additional Example 3 Continued

Let r represent the price of one rose.

39.95 + 2.95r = 26.00 + 4.50r

Subtract 2.95r from both sides.

– 2.95r– 2.95r

39.95 = 26.00 + 1.55r

Subtract 26.00 from both sides.

– 26.00– 26.00

13.95 = 1.55r

Divide both sides by 1.55.

9 = r

The two services would cost the same when purchasing 9 roses.

Additional Example 5: Solving Literal Equations for a Variable

The equation t = m + 10e gives the test score t for a student who answers m multiple-choice questions and e essay questions correctly. Solve this equation for e.

t = m + 10e

Locate e in the equation.

t = m + 10e

Since m is added to 10e, subtract m from both sides.

t – m = 10e

–m –m

Since e is multiplied 10, divide both sides by 10.

t – m = 10e

10

10

t – m = e

10

1

1

2

4

Lesson Quiz

Solve.

1. 4x + 16 = 2x

2. 8x – 3 = 15 + 5x

3. 2(3x + 11) = 6x + 4

4.x = x – 9

5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each?

x = –8

x = 6

no solution

x = 36

An orange has 45 calories. An apple has 75 calories.