Post on 08-Aug-2020
transcript
Literal Equations Scavenger hunt
Students should begin on an assigned problem. Upon completing that problem, they will find their answer on another card. Then, solve that problem. Continue
until all problems have been solved.
Solve for y: 4π₯ + 2π¦ = 6
Solve for y: β2π₯ + 2π¦ = 8
1
2
Answer: π = π + 2π + 6
Answer: π¦ = β
83π₯ β 2
Solve for y: 2π₯ β π¦3 = β2
Solve for y: β2(π¦ + 1) = 8π₯
3
4
Answer:
π = β23 π + 3
Answer: π¦ =
73π₯ β 7
Solve for y: β3π¦2 β 4π₯ = 3
Solve for y: 7π₯ β 3π¦ = 21
6
5
Answer:
π =βπ β 6π
Answer: π¦ = 2π₯ + 6
Solve for y: π₯ + π¦3 = 4
Solve for y: 3π¦ β 5π₯ = β10
7
8
Answer:
π¦ =53 π₯ β
103
Answer:
π =β3π + π
4
Solve for a: 3π + 4π = π
Solve for b: β2π β 3π = β9
9
10
Answer: π¦ = β4π₯ + 1
Answer: π¦ = βπ₯ + 12
Solve for a: ππ + π2 = β3
Solve for a: π β π2 = π + 3
11
12
Answer: π¦ = β2π₯ + 3
Answer: π¦ = π₯ + 4
Answers: 1 ! 11 ! 4 ! 9 ! 8 ! 7 ! 10 ! 3 ! 5 ! 2 ! 12 ! 1