Algebra I 3.1 Solving Equations. Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7.

Algebra I 3.1 Solving Equations

Solve an equation using subtraction

EXAMPLE 1

Solve x + 7 = 4.

x + 7 = 4 Write original equation.

x + 7 – 7 = 4 – 7 Subtract 7 from each side.

x = – 3 Simplify.

CHECK Substitute – 3 for x in the original equation.

x + 7 = 4 Write original equation.

–3 + 7 = 4? Substitute –3 for x.

4 = 4 Simplify. Solution checks.

Solve an equation using addition

EXAMPLE 2

Solve x – 12 = 3.

x – 12 = 3

+12 +12

x = 15

Solve an equation using division

EXAMPLE 3

Solve – 6x = 48.

– 6x = 48

x = – 8

Write original equation.

– 6x– 6

48– 6= Divide each side by – 6.

Simplify.

GUIDED PRACTICE for Example 2 and 3

1. y + 7 = 10

y + 7 = 10

y + 7 – 7 = 10 – 7

y = 3

Solve the equation. Check your solution.

2. x – 5 = 3

x – 5 = 3

+5 +5

x = 8

q – 11 = – 5

q = 6

+11– 11

3. q – 11 = – 5

GUIDED PRACTICE for Example 2 and 3

CHECK #1

y + 7 = 10

3 + 7 = 10?

10 = 10

CHECK #2

CHECK #3

8 – 5 = 3?

3 = 3

x – 5 = 3

6 – 11 = –5?

–5 = –5

q – 11 = –5

GUIDED PRACTICE for Example 2 and 3

4. 6 = t – 2.

6 = t – 2

+ 2 + 2

8 = t

Solve the equation. Check your solution.

4x = 48

x = 12

4x 4

48 4=

5. 4x = 48.

CHECK #4 CHECK #5

6 = 8 – 2?

6 = 6

6 = t – 2 4x = 48.

48 = 48

4 12 = 48?

GUIDED PRACTICE for Example 2 and 3

6. – 65 = – 5y.

– 65 = – 5y

13 = y

– 65– 5

– 5y– 5=

Solve the equation. Check your solution.

6w = – 54

w = – 9

6w 6

– 54 6=

7. 6w = – 54.

– 65 = – 5y

– 65 = – 65

– 65 = – 5? 13

6w = – 54

– 54 = – 54

6 – 9 = – 54?

CHECK #6 CHECK #7

GUIDED PRACTICE for Example 2 and 3

8. 24 = – 8n.

24 = – 8n

– 3 = n

24– 8

– 8n– 8=

Solve the equation. Check your solution.

24 = – 8n

24 = 24

24 = – 8? – 3

CHECK #8

Solve an equation using multiplicationEXAMPLE 4

Write original equation.

4 x4 = 4 5 Multiply each side by 4.

x = 20 Simplify.

= 5x 4Solve

SOLUTION

= 5x4

GUIDED PRACTICE for Example 4

SOLUTION

9. = 9t– 3

= 9t– 3

Write original equation.

– 3 t– 3 = – 3 9 Multiply each side by –3.

t = –27 Simplify.

Solve the equation. Check your Solution.

9 = 9

= 9.t– 3

= 9.– 27– 3

?

CHECK #9

GUIDED PRACTICE for Example 4

SOLUTION

Write original equation.

= 7 7 c 7

6 Multiply each side by 7.

42 = c Simplify.

10. 6 c7=

6 c7=

Solve the equation. Check your Solution.

CHECK #10

6 = 6

c 7=6

?= 42 7

6

GUIDED PRACTICE for Example 4

SOLUTION

Write original equation.

= – 2 – 2 z – 2 13 Multiply each side by – 2.

– 26 = z Simplify.

11. 13 z– 2=

– 2 13 =z

Solve the equation. Check your Solution.

13 = 13

z – 2=13

?= – 26 – 2

13

CHECK #11

GUIDED PRACTICE for Example 4

12. = – 11a 5

Write original equation.

5a 5 = 5 – 11 Multiply each side by 5.

a = – 55 Simplify.

= – 11a 5

Solve the equation. Check your Solution.

– 11 = – 11

= – 11a 5

= – 11 – 555

?

CHECK #12

Solve an equation by multiplying by a reciprocal

EXAMPLE 5

Write original equation.x = 427

–

x = 427

–Solve 72–The coefficient of x is

72

–The reciprocal of27

–is

( )(x ) =27

–72

–72

– 4Multiply each side by the

72

–reciprocal,

x = – 14 Simplify.

Solve an equation by multiplying by a reciprocal

EXAMPLE 5

ANSWER

The solution is – 14. Check by substituting – 14 for x in the original equation.

CHECK Write original equation.

Substitute –14 for x.

4 = 4 Simplify. Solution checks.

x = 427

–

(–14) = 4 ?2

7–

GUIDED PRACTICE for Example 5

Solve the equation. Check your Solution.

Write original equation.

w = 105613.

5 ( )(w ) =56

6 65 10

5

Multiply each side by the 6reciprocal,

65The coefficient of w is

565

6The reciprocal of is

w = 1056

w = 12 Simplify.

10 = 10

w = 1056

(12) = 10?5

6

CHECK #13

GUIDED PRACTICE for Example 5

Solve the equation. Check your Solution.

Write original equation.

p = 142314.

2 ( )(p ) =23

3 32 14

2

Multiply each side by the 3reciprocal,

p = 1423

32

.The coefficient of p is

232

3 is .The reciprocal of

p = 21 Simplify.

14 = 14

p = 1423

(21) = 14?2

3

CHECK #14

GUIDED PRACTICE for Example 5

Solve the equation. Check your Solution.

Write original equation.

3

Multiply each side by the – 4reciprocal,

.4– 3The coefficient of m is

3– 4– 3

4The reciprocal of is .

15. 9 = – 3

4m

9 = – 3

4m

– 3(– 4

3 9 ) 3 ( m ) 4

– 4=

– 12 = m Simplify.

9 = 9

9 = – 3

4m

49 =

– 3 (12)?

CHECK #15

GUIDED PRACTICE for Example 5

Solve the equation. Check your Solution.

Write original equation.

4

Multiply each side by the – 5reciprocal,

16. – 8 = – 4

5v

– 4–(

– 54 8) 4 ( v )

5– 5

=

– 8 = – 4

5v

4– 5– 4

5The reciprocal of is .

5– 4The coefficient of v is .

10 = v Simplify.

–8 = – 8

– 8 = – 4

5v

58 =

– 4 (10)?–

CHECK #16

EXAMPLE 6 Write and solve an equation

Let r represent Crawford's speed in meters per second. Write a verbal model. Then write and solve an equation.

OLYMPICS

SOLUTION

In the 2004 Olympics, Shawn Crawford won the 200 meter dash.

His winning time was 19.79 seconds. Find his average speed to the nearest tenth of a meter per second.

Write and solve an equation

EXAMPLE 6

Crawford's average speed was about 10.1 meters per second.

ANSWER

19.7920019.79 19.79 r=

200 = r 19.79

10.1 r

GUIDED PRACTICE for Example 6

17. WHAT IF? In the example 6, suppose Shawn Crawford ran 100 meters at the same average speed he ran the 200 meters. How long would it take him to run 100 meters ? Round your answer to nearest tenth of a second.

Hint: Let t represent Crawford’s time in speed.

100 = 10.1 t

10.1t10010.1 10.1

=

9.9 t

Therefore, about 9.9 seconds

required Crawford’s to run

100 meters.