if you x A by 7 and B by 3 the sum has a diff. of 12, 3/4 of A and 2/3 of B=21 solve for A&B

Question

This was a question i was asked a while ago and for some reason cant wrap my brain around it.

Answer 1

if you x A by 7 and B by 3 the sum has a diff. of 12, 3/4 of A and 2/3 of B=21 solve for A&B

1) 7A + 3B = 12        Multiply by 8

1a) 56A + 24B = 96

2) 3/4A + 2/3B = 21  Multiply by 12

2a) 9A + 8B = 252    Multiply by 3

2b) 27A + 24B = 756 Subtract 1a) from 2b)

1a) 56A + 24B = 96

--------------------------------

     -29A = 660

          A = 660/-29 = -22.76

1) 7A + 3B = 12  plug in A = -22.76

7(-22.76) + 3B = 12

-159.32 + 3B = 12

3B = 171.32

B = 57.11

A= -22.76, B = 57.11