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
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
