Worked Example/algebra

Question

Luis has 58 coins in nickels, dimes, and quarters. The number of nickels is three less than twice the number of dimes. The total value of the coins is $7.40. How many of each type of coin does Luis have?

Answer 1

Using the letters Q, N and D for the numbers of quarters nickels and dimes we can write:

Q+N+D=58 coins.

25Q+5N+10D=740 cents.

N=2D-3.

So Q+2D-3+D=58, Q=61-3D. And 25(61-3D)+5(2D-3)+10D=740.

Therefore, 1525-75D+10D-15+10D=740.

1510-55D=740, divide through by 5: 302-11D=148, 11D=154, so D=14, N=28-3=25, Q=61-42=19.

A quick check shows there are 58 coins=19+25+14, value is 4.75+1.25+1.40=$7.40.