Coin Problem

Harold has $4.35 of change in quarters and dimes, with 5 more dimes than quarters. How many of each coin does he have?  

x = number of quarters
x + 5 = number of dimes   {there are 5 more dimes than quarters} 

25x + 10(x + 5) = 435   {number of coins times value of coin equals total value}
25x + 10x + 50 = 435   {used distributive property}
35x + 50 = 435   {combined like terms}
35x = 385   {subtracted 50 from each side}
x = 11   {divided each side by 35}
x + 5 = 16   {substituted 11, in for x, into x + 5} 

11 quarters and 16 dimes

- Algebra House