50 f + 20 t = 310
f + t = 11 ... 20 f + 20 t = 220
subtracting equations (to eliminate t)
30 f = 90 ... f = 3
substitute back to find t
Q) Julie has $3.10 in change in her pocket. If she has only 50 cent and 20 cent pieces and the total number of coins is 11, how many coins of each type does she have?
2 answers
50x + 20y = 310
x + y = 11
multiply bottom equation by 20
50x + 20y = 310
20x + 20y = 220
substitute into 1st equation via elimination
30x = 90
x = 90 รท 30
x = 3
solve for y
3 + y = 11
y = 11 - 3
y = 8
Answer: 3 50c pieces and 8 20c pieces
x + y = 11
multiply bottom equation by 20
50x + 20y = 310
20x + 20y = 220
substitute into 1st equation via elimination
30x = 90
x = 90 รท 30
x = 3
solve for y
3 + y = 11
y = 11 - 3
y = 8
Answer: 3 50c pieces and 8 20c pieces