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?

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