Find two numbers such that five times the larger plus three times the smaller is 47, and four times the larger minus twice the smaller is 20.

Let x and y be the larger and smaller numbers respectively.
Use the given fact "five times the larger plus three times the smaller is 47" to set up an equation:
5x + 3y = 47 (1)
Use the given fact "four times the larger minus twice the smaller is 20" to set up another equation:
4x - 2y = 20 (2)
Multiply both sides of (1) by 2, we have:
10x + 6y = 94 (3)
Multiply both sides of (2) by 3, we have:
12x - 6y = 60 (4)
Add (3) and (4), we have
22x = 154
x = 7
Substituting x = 7 into (1), we have
5*7 + 3y = 47
Solving for y, we have:
35 + 3y = 47
3y = 12
y = 4

Ibelieve the answer is c