Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 Course 3...

Warm UpSolve.

1. 2x + 9x – 3x + 8 = 16

2. –4 = 6x + 22 – 4x

3. + = 5

4. – = 3

Course 3

11-3 Solving Equations with Variables on Both Sides

27

x7 7

1

9x16

2x4

18

Learn to solve equations with variables on both sides of the equal sign. Determine if the equation has a solution and the type of solution.

Course 3


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.

Course 3


Solve.

4x + 6 = x

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

4x + 6 = x– 4x – 4x

6 = –3x

Subtract 4x from both sides.

Divide both sides by –3.

–2 = x

6–3

–3x–3=

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Course 3


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

Helpful Hint

Solve.

9b – 6 = 5b + 18

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

9b – 6 = 5b + 18– 5b – 5b

4b – 6 = 18

4b 4

24 4 =

Subtract 5b from both sides.

Divide both sides by 4.

b = 6

+ 6 + 6

4b = 24Add 6 to both sides.

Course 3


Solve.

9w + 3 = 9w + 7

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

3 ≠ 7

9w + 3 = 9w + 7

– 9w – 9w Subtract 9w from both sides.

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

Course 3


Course 3


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

Helpful Hint

Solve.

5x + 8 = x

Check It Out: Example 1A

5x + 8 = x– 5x – 5x

8 = –4x

Subtract 5x from both sides.

Divide both sides by –4.

–2 = x

8–4

–4x–4=

Course 3


Solve.

3b – 2 = 2b + 123b – 2 = 2b + 12

– 2b – 2b

b – 2 = 12

Subtract 2b from both sides.

+ 2 + 2

b = 14Add 2 to both sides.

Check It Out: Example 1B

Course 3


Solve.

3w + 1 = 3w + 8

1 ≠ 8

3w + 1 = 3w + 8

– 3w – 3w Subtract 3w from both sides.

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

Check It Out: Example 1C

Course 3


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.

Course 3


Solve.

10z – 15 – 4z = 8 – 2z - 15

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

10z – 15 – 4z = 8 – 2z – 15

+ 15 +15

6z – 15 = –2z – 7 Combine like terms.+ 2z + 2z Add 2z to both sides.

8z – 15 = – 7

8z = 8

z = 1

Add 15 to both sides.

Divide both sides by 8.8z 88 8=

Course 3


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

Multiply by the LCD, 20.

4y + 12y – 15 = 20y – 14

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

y5

34

3y5

710

+ – = y –

y5

34

3y5

710

+ – = y –

20( ) = 20( )y5

34

3y5

710

+ – y –

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

3y5

34

710

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Additional Example 2B Continued


–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.

Course 3


Solve.

12z – 12 – 4z = 6 – 2z + 32

Check It Out: Example 2A

12z – 12 – 4z = 6 – 2z + 32

+ 12 +12

8z – 12 = –2z + 38 Combine like terms.+ 2z + 2z Add 2z to both sides.

10z – 12 = 38

10z = 50

z = 5


Divide both sides by 10.10z 5010 10=

Course 3


Multiply by the LCD, 24.

6y + 20y + 18 = 24y – 18

26y + 18 = 24y – 18 Combine like terms.

y4

34

5y6

68

+ + = y –

y4

34

5y6

68

+ + = y –

24( ) = 24( )y4

34

5y6

68

+ + y –

24( ) + 24( )+ 24( )= 24(y) – 24( )y4

5y6

34

68

Check It Out: Example 2B

Course 3


Subtract 18 from both sides.

2y + 18 = – 18

2y = –36

– 18 – 18

–36 2

2y2 = Divide both sides by 2.

y = –18

26y + 18 = 24y – 18

– 24y – 24y Subtract 24y from both sides.

Check It Out: Example 2B Continued

Course 3


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.

Course 3


Additional Example 3 Continued

39.95 + 2.95r = 26.00 + 4.50rLet r represent the price of one rose.

– 2.95r – 2.95r

39.95 = 26.00 + 1.55r

Subtract 2.95r from both sides.

– 26.00 – 26.00 Subtract 26.00 from both sides.

13.95 = 1.55r 13.951.55

1.55r 1.55= Divide both sides by 1.55.

9 = rThe two services would cost the same when purchasing 9 roses.

Course 3


Check It Out: Example 3

Marla’s Gift Baskets sell a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both gift basket companies muffin baskets cost the same price.

Course 3


Check It Out: Example 3 Continued

22.00 + 2.25b = 16.00 + 3.00bLet b represent the price of one balloon.

– 2.25b – 2.25b

22.00 = 16.00 + 0.75b

Subtract 2.25b from both sides.

– 16.00 – 16.00 Subtract 16.00 from both sides.

6.00 = 0.75b 6.00.75

0.75b 0.75

= Divide both sides by 0.75.

8 = bThe two services would cost the same when purchasing 8 balloons.

Course 3


Additional Example 4: Multi-Step 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?

Course 3



First solve for the price of one doughnut.

1.25 + 2d = 0.50 + 5dLet d represent the price of one doughnut.

– 2d – 2d

1.25 = 0.50 + 3dSubtract 2d from both sides.

– 0.50 – 0.50Subtract 0.50 from both sides.

0.75 = 3d

0.753

3d3= Divide both sides by 3.

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

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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.

Course 3


Check It Out: Example 4

Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

Course 3



First solve for distance around the track.

2x + 4 = 4x + 2Let x represent the distance around the track.

– 2x – 2x

4 = 2x + 2Subtract 2x from both sides.

– 2 – 2 Subtract 2 from both sides.

2 = 2x

22

2x2= Divide both sides by 2.

1 = x The track is 1 mile around.

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Now find the total distance Helene walks each day.

2x + 4 Choose one of the original expressions.

Helene walks 6 miles each day.2(1) + 4 = 6

Let n represent the number of 1-mile laps.

Find the number of laps Helene walks on Saturdays.

1n = 6

Helene walks 6 laps on Saturdays.

n = 6

Course 3


Try This!

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?

Insert Lesson Title Here

14

12

Course 3


Homework:Workbook pg. 85

Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 Course 3...

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Transcript of Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 Course 3...