Question

Steve had 7 more than twice as many quarters as dimes. if the total value of her coins was $10.15, how many of each kind of coin did she have?

Answer 1

x=quarters, y=dimes
7+2x=y
0.25x+0.1y=10.15
we substitute the first equation into the second equation
0.25x+0.1(7+2x)=10.15
0.25x+0.7+0.2x=10.15
0.45x+0.7-0.7=10.15-0.7
0.45x=9.45
0.45x/0.45=9.45/0.45
x=21 quarters
7+2x=y, x=21
7+2(21)=y
7+42=y
y=49 dimes
check: 0.25x+0.1y=10.15 when x=21, y=49
21(0.25)=5.25 in quarters
49(0.1)=4.90 in dimes
5.25+4.90=10.15