Lesson Review

Directions: Solve the following equations by using the rules for removing the parentheses.

1. 2(x - 3) = 32

2. 6(2x + 4) = 24

3. 3(3s + 4) = 21

4. 6(c - 2) = 4(c + 3)

5. 12(r - 3) = 6(r + 3)

6. 3(x + 5) = 30

7. 6(b + 2) = 3(b-2)

8. 2(c - 2) = 16

9. 4(y + 2) = 28

10. 2(d - 2) = 2(d + 3)

11. 5(L + 5) = 35

12. 2(k - 2) = 44

13. p - 9 = 11

14. 3(s + 3) = 27

15. 5(a - 3) = 45

1.) x= 19

2.)x= 0

3.)s= 1

4.)c= 12

5.)r= 9

6.)x= 5

7.)b= -6

8.)c= 10

9.)y= 5

10.) no real solution

11.)L= 2

12.)k= 24

13.)p= 20

14.)s= 6

15.)a= 12

Step-by-step explanation:

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.