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Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front Hand in HW – p 80 to front table table 2-2 Notes, HW p 87 2-2 Notes, HW p 87 # 21-44, 101-104 # 21-44, 101-104

Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

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Page 1: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing2-2 Solving Equations by

Multiplying or Dividing

Holt Algebra 1

Hand in HW – p 80 to front tableHand in HW – p 80 to front table

2-2 Notes, HW p 872-2 Notes, HW p 87

# 21-44, 101-104# 21-44, 101-104

Page 2: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

I will:

Solve one-step equations in one variable by using multiplication or division.

Objective

Page 3: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve = 5.

Solving Equations Using Multiplication

b–4b–4 = 5–4 • • –4 Multiply both sides by –4.

b = –20

Checkb–4= 5

Substitute –20 for b.–20–4 = 5

?

5 = 5?

2-2 Solving Equations by Multiplying or Dividing

Course 3

Think…. What type of problem is this and how do I “undo” it?

Page 4: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve = 5.

Check It Out: Example 2

c–3c–3 = 5–3 • • –3 Multiply both sides by –3.

c = –15

Checkc–3= 5

Substitute –15 for c.–15–3 = 5

?

5 = 5?

2-2 Solving Equations by Multiplying or Dividing

Course 3

Page 5: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Example 2A: Solving Equations by Using Division

Since y is multiplied by 9, divide both sides by 9 to undo the multiplication.y = 12

108 108

To check your solution, substitute 12 for y in the original equation.

9y = 108

9(12) 108

Check 9y = 108

Page 6: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Example 2B: Solving Equations by Using Division

Since v is multiplied by –6, divide both sides by –6 to undo the multiplication.

0.8 = v

–4.8 –4.8

To check your solution, substitute 0.8 for v in the original equation.

–4.8 = –6v

–4.8 –6(0.8)

Check –4.8 = –6v

Page 7: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Check It Out! Example 2a

Since c is multiplied by 4, divide both sides by 4 to undo the multiplication.

4 = c

16 16

To check your solution, substitute 4 for c in the original equation.

16 = 4c

16 4(4)

Check 16 = 4c

Page 8: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Check It Out! Example 2c

Since k is multiplied by 15, divide both sides by 15 to undo the multiplication.k = 5

75 75

To check your solution, substitute 5 for k in the original equation.

15k = 75

15(5) 75

Check 15k = 75

Page 9: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Remember that dividing is the same as multiplying by the reciprocal.

Instead of dividing by a fraction to get the variable alone, you multiply both sides by the reciprocal.

Page 10: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

What is the reciprocal of:

• 5/6

• 2

• -9/8

Page 11: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation.

Example 3A: Solving Equations That Contain Fractions

w = 24

20 20

To check your solution, substitute 24 for w in the original equation.

w = 2056

Check w = 2056

The reciprocal of is . Since w is

multiplied by , multiply both sides

by .

56

65

566

5

20

Page 12: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation.

Example 3B: Solving Equations That Contain Fractions

= z3

16

To check your solution,

substitute for z in the

original equation.

32

= z32

18

The reciprocal of is 8. Since z is

multiplied by , multiply both sides

by 8.

181

8

Check18

316

= z

316

316

Page 13: Holt Algebra 1 2-2 Solving Equations by Multiplying or Dividing 2-2 Solving Equations by Multiplying or Dividing Holt Algebra 1 Hand in HW – p 80 to front

Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Solve the equation. Check your answer.Check It Out! Example 3a

– = b14

To check your solution,

substitute – for b in the

original equation.

54

15

The reciprocal of is 5. Since b is

multiplied by , multiply both sides

by 5.

151

5

= b54–

= b5 4Check11–