Pre-Algebra 10-3 Solving Equations with Variables on Both Sides 10-3 Solving Equations with Variables on Both Sides Pre-Algebra HOMEWORK & Learning Goal.

Pre-Algebra

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

Variables on Both Sides

Pre-Algebra

HOMEWORK & Learning GoalHOMEWORK & Learning Goal

Lesson PresentationLesson Presentation

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

PA HOMEWORK Answers

Page 504

#12-30 EVENS

Pre-Algebra


Pre-Algebra HOMEWORK

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#10-28 EVENS

Pre-Algebra


Our Learning GoalStudents will be able to solve

multi-step equations with multiple variables, solve

inequalities and graph the solutions on a number line.

Pre-Algebra


Our Learning Goal Assignments• Learn to solve two-step equations.

• Learn to solve multistep equations.

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

• Learn to solve two-step inequalities and graph the solutions of an inequality on a number line.

• Learn to solve an equation for a variable.

• Learn to solve systems of equations.

Pre-Algebra


Today’s Learning Goal Assignment

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

Pre-Algebra


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.

Solving Strategy for Variables!

Pre-Algebra


Solve.

A. 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=

Pre-Algebra


Solve.

A. 5x + 8 = x

Try This: Example 1A

5x + 8 = x– 5x – 5x

8 = –4x

Subtract 4x from both sides.

Divide both sides by –4.

–2 = x

8–4

–4x–4=

Pre-Algebra


Solve.

B. 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.

Pre-Algebra


Solve.

B. 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.

Try This: Example 1B

Pre-Algebra


Solve.

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

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

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

3 ≠ 7

9w + 3 = 9w + 7 Combine like terms.

– 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.

Pre-Algebra


Solve.

C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w

1 ≠ 8

3w + 1 = 3w + 8 Combine like terms.

– 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.

Try This: Example 1C

Pre-Algebra


To solve multistep equations with variables on both sides:

1.Combine like terms and clear fractions

2.Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation

3.Then use properties of equality to isolate the variable.

How to tackle a beast of an equation!

Pre-Algebra


Solve.

A. 10z – 15 – 4z = 8 – 2z - 15

Additional Example 2A: Solving Multistep 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=

Pre-Algebra


Solve.

A. 12z – 12 – 4z = 6 – 2z + 32

Try This: 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=

Pre-Algebra


B.

Additional Example 2B: Solving Multistep Equations with Variables on Both Sides

Multiply by the LCD.

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

Pre-Algebra


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.

Pre-Algebra


B.

Multiply by the LCD.

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

Try This: Example 2B

Pre-Algebra


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.

Try This: Example 2B Continued

Pre-Algebra


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


Additional Example 3 Continued

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.

Pre-Algebra


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.

Pre-Algebra


Try This: Example 3

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?

Pre-Algebra


Try This: Example 3 Continued

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.

Pre-Algebra


Try This: Example 3 Continued

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

Pre-Algebra


Don’t forget your proper heading! Trade & Grade!

10-3 Lesson QuizSolve.

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 = 6

x = –8

no solution

x = 3614

12

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

Pre-Algebra 10-3 Solving Equations with Variables on Both Sides 10-3 Solving Equations with Variables on Both Sides Pre-Algebra HOMEWORK & Learning Goal.

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