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.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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.       Log On


   



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) About Me  (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