SOLUTION: Gerald had $1.70 in dimes and quaters. He had 3 more dimes than quaters. How's many of each coin did he have.
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Question 849675: Gerald had $1.70 in dimes and quaters. He had 3 more dimes than quaters. How's many of each coin did he have.
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
.10d + .25q = 1.70
d = q + 3
.10(q+3) + .25q = 1.70
.10q + .30 + .25q = 1.70
.35q = 1.40
q = 4
d = 7
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