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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