college algebra word problem

Question

A newspaper carrier has $4.85 in change. He has ten more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he​ have?

Answer 1

Break the problem down into its components:

Let X=number of dimes

Let X+10= number of quarters

Let 2*(X+10) = number of nickels

Total amount = $4.85

X*.10+(X+10)*.25+2*(x+10)*.05 = $4.85 total

Expand the equation:

.10X +.25X + 2.5+(2x+20)*.05=4.85

.10X +.25X+ .10X+ 1+2.5=4.85

.45X +3.5=4.85

Solve for X:

.45X=4.85-3.5

.45X=1.35

X=3

So...

There are 3 dimes; 13 quarters; and, 26 nickels.

$4.85 total