Problema Solution
Tyler had $2.90 in nickels, dimes, and quarters. He had two more dimes than quarters, and he had twice as many nickels as quarters. How many of each did he have?
Answer provided by our tutors
1 nickle = 5 cents
1 dime = 10 cents
1 quarter = 25 cents
let
n = the number of nickels
d = the number of dimes
q = the number of quarters
Tyler had $2.90 = 290 cents in nickels, dimes, and quarters
5n + 10d + 25q = 290 divide both sides by 5
n + 2d + 5q = 58
He had two more dimes than quarters
d = q + 2
he had twice as many nickels as quarters
n = 2q
by solving the system of equations
n + 2d + 5q = 58
d = q + 2
n = 2q
we find
n =12 nickles
d = 8 dimes
q = 6 quarters
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