Question

Arnold had $1.70 in dimes and quarters. He had 3 more dimes thing quarters. How many of each coin did he have?

Answer 1

$x-number\ of\ quarters\\x+3-number\ of\ dimes\\\\0.25x+0.1(x+3)=1.7\\0.25x+0.1x+0.3=1.7\\0.25x+0.1x=1.7-0.3\\0.35x=1.4\ \ \ \ \ |:0.35\\x=4\\\\x+3=4+3=7\\\\He\ had\ 4\ quarters\ and\ 7\ dimes.$