Fred had a sum of money. If he bought 7 shortbreads and 8 wafers, he would be short of $6. If he bought 2 shortbreads and 4 wafers, he would have $16 left. Given that he had $32, how many wafers could he buy at most?
Cost of 1 shortbread = 1 S
Cost of 1 wafer = 1 W

7 S + 8 W - 6 = 32
7 S + 8 W = 32 + 6 
7 S + 8 W = 38 --- (1)

2 S + 4 W + 16 = 32 
2 S + 4 W = 32 - 16 
2 S + 4 W = 16 --- (2)

Make S the same.

(1) _{x 2}
14 S + 16 W = 76 
14 S = 76 - 16 W ---- (3)

(2) _{x 7}
14 S + 28 W = 112 
14 S = 112 - 28 W ---- (4)

(3) = (4)
76 - 16 W = 112 - 28 W 
28 W - 16 W = 112 - 76 
12 W = 36

1 W = 36 ÷ 12 = 3 

Cost of 1 wafer = $3 

Number of wafers that Fred can buy at most
= 32 ÷ 3
= 10 r 2

Most number of wafers that Fred can buy = 10

Answer(s): 10