HW # 32- Videos for section 3-4 AND p. 132 # 1-11 all Website: go.hrw.com keyword: MT8CA 3-4 Warm up

a) Lydia rode 243 miles in a three-day bike trip. On the first day, Lydia rode 67 miles. On the second day, she rode 92 miles. How many miles per hour did she average on the third day if she rode for 7 hours?

http://www.mvschools.org/Page/2248

Week 9, Day Two

Warm Up Response

a) n=3b) k= -10c) a=4d) Lydia rode 243 miles in a three-day bike trip. On the

first day, Lydia rode 67 miles. On the second day, she rode 92 miles. How many miles per hour did she average on the third day if she rode for 7 hours?

12 miles/hour

Homework Videos

Homework Checkp. 126 # 12-23 all AND p. 127 # 41-4812) n=313) k=-1014)c=215)w=316)a=417)y=518)p=2219) h=620)g=321) m=2

41) C42) 9243) n=-344) y=-245) x=12146) 1547) 6t+3k-1548) 5a-b+4

22) b=423) x=-12

p. 126 # 12-23 all AND p. 127 # 41-48

20.) 22.)

Goals for Today

• Solving Equations with Variables on Both Sides

(3-4) {AltOpener 3-4 Solv Eq w Variab on Both Sides.pdf}

• Simplifying Expressions #3 Worksheet

Slides for extra practice at home

• Note, some of the formatting may be off and the ppt moves from Mac to PC.

Solve.

4x + 6 = x

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

4x + 6 = x

– 4x – 4x

6 = –3x

To collect the variable terms on one side, subtract 4x from both sides.

Since x is multiplied by -3, divide both sides by –3.

–2 = x

6–3

–3x–3

=

You can always check your solution by substituting the value back into the original equation.

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

=

To collect the variable terms on one side, subtract 5b from both sides.

Since b is multiplied by 4, divide both sides by 4.

b = 6

+ 6 + 6

4b = 24

Since 6 is subtracted from 4b, add 6 to both sides.

Solve.

9w + 3 = 9w + 7

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

3 ≠ 7

9w + 3 = 9w + 7

– 9w – 9w To collect the variable terms on one side, subtract 9w from both sides.

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

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

Since x is multiplied by –4, divide both sides by –4.

–2 = x

8–4

–4x–4

=

To collect the variable terms on one side, subtract 5x from both sides.

Solve.

3b – 2 = 2b + 12

3b – 2 = 2b + 12

– 2b – 2b

b – 2 = 12

+ 2 + 2

b = 14

Since 2 is subtracted from b, add 2 to both sides.

Check It Out! Example 1B

To collect the variable terms on one side, subtract 2b from both sides.

Solve.

3w + 1 = 3w + 8

1 ≠ 8

3w + 1 = 3w + 8

– 3w – 3w To collect the variable terms on one side, 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

To solve more complicated equations, you may need to first simplify by combining like terms or clearing fractions. Then add or subtract to collect variable terms on one side of the equation. Finally, use properties of equality to isolate the variable.

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

=

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

Additional Example 2B Continued

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.

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

Add 12 to both sides.

Divide both sides by 10.10z 5010 10

=

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

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

Additional Example 3: Business Application

Daisy’s Flowers sells 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. What is the price?

Daisy’s: c = 39.95 + 2.95 r

Write an equation for each service. Let c represent the total cost and r represent the number of roses.

total cost is flat fee plus cost for each rose

Other: c = 26.00 + 4.50 r

Additional Example 3 Continued

39.95 + 2.95r = 26.00 + 4.50rNow write an equation showing that the costs are equal.

– 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 bouquets from either florist would cost the same when purchasing 9 roses.

Additional Example 3 Continued

To find the cost, substitute 9 for r into either equation.

Daisy’s:

The cost for a bouquet with 9 roses at either florist is $66.50.

c = 39.95 + 2.95r

c = 39.95 + 2.95(9)

c = 39.95 + 26.55

c = 66.5

Other florist:

c = 26.00 + 4.50r

c = 26.00 + 4.50(9)

c = 26.00 + 40.50

c = 66.5

Check It Out! Example 3Marla’s Gift Baskets sells 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 baskets cost the same price.

Marla’s: c = 22.00 + 2.25 b

total cost is flat fee plus cost for each balloon

Other: c = 16.00 + 3.00 b

Write an equation for each service. Let c represent the total cost and b represent the number of balloons.

Check It Out! Example 3 Continued

22.00 + 2.25b = 16.00 + 3.00bNow write an equation showing that the costs are equal.

– 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

0.750.75b 0.75= Divide both sides by 0.75.

8 = bThe two services would cost the same when purchasing a muffin basket with 8 balloons.