*Math 1 Foundations Jeopardy*

Solve for x

100

11 = 4x-13

x = 6

200

3(x+6) = 24

x = 2

300

8(x+2) = 32

x = 2

400

18 - 4m = -10 - 6m

x = -4

500

−5(1 − 5x) + 5(−8x − 2) = −4x − 8x

x=-5

Solve for x Pt. 2

100

6(2 + x) + 6 = 24

x = 1

200

5x - 13x +2x = -60 

x = 10

300

8k + 72 = 32

x = 5

400

4 (2x + 5) + 4 = 3(5x - 6)

x = 6

500

-8x + 5 + 3x = -10x -20

x = 5

Combine like Terms

100

9 + a - 5b -2

a - 5b +7

200

6 + 2a + 10

16 + 2a

300

7k - 4k + 13

3k +13

400

10m + 7 + 3m - 9

13m - 2 

500

8x - 5y + 9 + 3x - 6

11x - 5y + 3

Distribute

100

4(6 - n)

24 - 4n 

200

2(5x - 3)

10x - 6

300

9(3 + 4z)

27 + 36z

400

15(2b + 3)

30b + 45

500

4(x - 5y + 7)

4x - 20y - 28

Types of Solutions

100

 7(2 + 5v) = 3v + 14

one solution

x=0

200

3 + 5n = 5(n + 2) - 7

Infinitely many solution

300

-5x+2-2(3+2x)=11x

One solution

-3=x

400

24a − 22 = −4(1 − 6a)

No solution

500

3(x+8)=15x-6+3x

one solution

x=2