solving equations with variables on both sides l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Solving Equations with Variables on Both Sides PowerPoint Presentation
Download Presentation
Solving Equations with Variables on Both Sides

Loading in 2 Seconds...

play fullscreen
1 / 24

Solving Equations with Variables on Both Sides - PowerPoint PPT Presentation


  • 90 Views
  • Uploaded on

Solving Equations with Variables on Both Sides. 10-3. Pre-Algebra. 1. 7. 1. 2 x. 9 x. 8. 16. 4. x. 2. 7. 7. Warm Up Solve. 1. 2 x + 9 x – 3 x + 8 = 16 2. – 4 = 6 x + 22 – 4 x 3. + = 5 4. – = 3. x = 1. x = -13. x = 34. x = 50.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Solving Equations with Variables on Both Sides' - elke


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide2

1

7

1

2x

9x

8

16

4

x

2

7

7

Warm Up

Solve.

1.2x + 9x – 3x + 8 = 16

2.–4 = 6x + 22 – 4x

3. + = 5

4. – = 3

x = 1

x = -13

x = 34

x = 50

slide4

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.

slide5

–3x

=

–3

6

–3

Example: Solving Equations with Variables on Both Sides

Solve.

A. 4x + 6 = x

4x + 6 = x

– 4x– 4x

Subtract 4x from both sides.

6 = –3x

Divide both sides by –3.

–2 = x

slide6

4b

24

=

4

4

Example: Solving Equations with Variables on Both Sides

Solve.

B. 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

slide7

9w + 3 = 9w + 7

Combine like terms.

– 9w– 9w

Subtract 9w from both sides.

Example: Solving Equations with Variables on Both Sides

Solve.

C. 9w + 3 = 5w + 7 + 4w

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

3 ≠ 7

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

slide8

–4x

=

–4

8

–4

Try This

Solve.

A. 5x + 8 = x

5x + 8 = x

– 5x– 5x

Subtract 4x from both sides.

8 = –4x

Divide both sides by –4.

–2 = x

slide9

Try This

Solve.

B. 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

slide10

3w + 1 = 3w + 8

Combine like terms.

– 3w– 3w

Subtract 3w from both sides.

Try This

Solve.

C. 3w + 1 = 10w + 8 – 7w

3w + 1 = 10w + 8 – 7w

1 ≠ 8

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

slide11

To solve multistep 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.

slide12

8z8

=

8

8

Example: Solving Multistep Equations with Variables on Both Sides

Solve.

A. 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

slide13

7

10

7

10

7

10

7

10

3y

5

3y

5

3y

5

3y

5

y

5

y

5

y

5

y

5

3

4

3

4

3

4

3

4

+ – = y –

20( ) = 20( )

+ – y –

20() + 20( ) – 20( )= 20(y) – 20( )

Example: Solving Multistep Equations with Variables on Both Sides

B.

+ – = y –

Multiply by the LCD.

4y + 12y – 15 = 20y – 14

16y – 15 = 20y – 14

Combine like terms.

slide14

4y

4

–1

-1

4

= y

=

4

Example Continued

16y – 15 = 20y – 14

– 16y– 16y

Subtract 16y from both sides.

–15 = 4y – 14

+ 14+ 14

Add 14 to both sides.

–1 = 4y

Divide both sides by 4.

slide15

10z50

=

10

10

Try This

Solve.

A. 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

slide16

6

8

6

8

6

8

6

8

5y

6

5y

6

5y

6

5y

6

y

4

y

4

y

4

y

4

3

4

3

4

3

4

3

4

+ + = y –

24( ) = 24( )

+ + y –

24() + 24( )+ 24( )= 24(y) – 24( )

Try This

B.

+ + = y –

Multiply by the LCD.

6y + 20y + 18 = 24y – 18

26y + 18 = 24y – 18

Combine like terms.

slide17

2y

2

–36

2

=

Try This Continued

26y + 18 = 24y – 18

– 24y– 24y

Subtract 24y from both sides.

2y + 18 = – 18

– 18– 18

Subtract 18 from both sides.

2y = –36

Divide both sides by 2.

y = –18

slide18

Example: 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?

slide19

=

3d

3

0.75

3

Example Continued

First solve for the price of one doughnut.

Let d represent the price of one doughnut.

1.25 + 2d = 0.50 + 5d

– 2d– 2d

Subtract 2d from both sides.

1.25 = 0.50 + 3d

Subtract 0.50 from both sides.

– 0.50– 0.50

0.75 = 3d

Divide both sides by 3.

The price of one doughnut is $0.25.

0.25 = d

slide20

1.75

0.25

=

0.25n

0.25

Example Continued

Now find the amount of money Jamie spends each morning.

Choose one of the original expressions.

1.25 + 2d

1.25 + 2(0.25) = 1.75

Jamie spends $1.75 each morning.

Find the number of doughnuts Jamie buys on Tuesday.

Let n represent the number of doughnuts.

0.25n = 1.75

Divide both sides by 0.25.

n = 7; Jamie bought 7 doughnuts on Tuesday.

slide21

Try This

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?

slide22

2

=

2

2x

2

Try This Continued

First solve for distance around the track.

Let x represent the distance around the track.

2x + 4 = 4x + 2

– 2x– 2x

Subtract 2x from both sides.

4 = 2x + 2

– 2– 2

Subtract 2 from both sides.

2 = 2x

Divide both sides by 2.

The track is 1 mile around.

1 = x

slide23

Try This Continued

Now find the total distance Helene walks each day.

Choose one of the original expressions.

2x + 4

2(1) + 4 = 6

Helene walks 6 miles each day.

Find the number of laps Helene walks on Saturdays.

Let n represent the number of 1-mile laps.

1n = 6

n = 6

Helene walks 6 laps on Saturdays.

slide24

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.