Question

Tom has $5.05 in quarters and dimes. How many coins are quarters if the total number of coins is 28?  Try creating a system of equations and solving this by either substitution or elimination.

Answer 1

Define the system of equations that describes the situation.

Take x as the number of quarters and y as the number of dimes, according to the question the total number of coins, which is the sum of quarter and dimes, is 28:

$x+y=28$

Also, using the equivalence in cents we know that each quarter is 25 cents and each dime is 10 cents. Expressed as dollars, a quarter equals 0.25 dollars and a dime equals 0.10 dollars. According to the question, the total amount of money is $5.05, this is the sum of the product of 0.25 and the number of quarters and the product of 0.10 and the number of dimes.:

$0.25x+0.10y=5.05$

Use both equations to define the system of equations:

$\begin{gathered} x+y=28 \\ 0.25x+0.10y=5.05 \end{gathered}$

Use substitution to solve the system:

$\begin{gathered} x+y=28 \\ y=28-x \end{gathered}$$\begin{gathered} 0.25x+0.1y=5.05 \\ 0.25x+0.1(28-x)=5.05 \\ 0.25x+2.8-0.1x=5.05 \\ 0.15x=5.05-2.8 \\ 0.15x=2.25 \\ x=\frac{2.25}{0.15} \\ x=15 \end{gathered}$

15 of the coins are quarters.