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 Comments are closed.

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