William has six more nickels than dimes. Combined, the total value of the coins is $1.35. How many of each does he have?
Let the number of dimes = N
Then the number of nickels = N + 6
So
number of dimes * value of each + number of nickels * value of each = total money
So we have
N * .10 + (N + 6) * .05 = 1.35 simplify
.10N + 05N + .30 = 1.35
.15N + .30 = 1.35 subtract .30 from both sides
.15N = 1.05 divide both sides by .15
N = 7 (dimes)
And the number of nickels = N + 6 = 7 + 6 = 13
Let the number of dimes = N
Then the number of nickels = N + 6
So
number of dimes * value of each + number of nickels * value of each = total money
So we have
N * .10 + (N + 6) * .05 = 1.35 simplify
.10N + 05N + .30 = 1.35
.15N + .30 = 1.35 subtract .30 from both sides
.15N = 1.05 divide both sides by .15
N = 7 (dimes)
And the number of nickels = N + 6 = 7 + 6 = 13