Q. 134 - Intermediate Algebra

Question

In the following exercise, translate to a system of equations and solve.

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value $$13.10$. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

Step-by-Step Solution

Answer

Peter has $48$ number of quarters and $11$ number of dimes.

Step 1. Given Information

The total value of quarters and dimes is $$13.10$.

Also, the number of quarters was fifteen more than three times the number of dimes. 

Step 2. Identify and name what we are looking for

We need to find the number of quarters and dimes Peter has.

Let $x$ represents the number of quarters and $y$ represents the number of dimes.

Step 3. Form the equations

The value of each quarter is $$0.25$ and the value of each dime is $$0.10$.

The value of $x$ quarters and $y$ dimes is $$13.10$. So an equation can be written as

$0.25x+0.10y=13.10 ...\left(1\right)$

The number of quarters was fifteen more than three times the number of dimes, so

$x=3y+15 ...\left(2\right)$

Step 4. Solve using substitution

Using the second equation, substitute $3y+15$ for $x$ in the first equation and solve for $y$

$\begin{array}{rcl}0.25x+0.10y& =& 13.10\\ 0.25(3y+15)+0.10y& =& 13.10\\ 0.75y+3.75+0.10y& =& 13.10\\ 0.75y+0.10y+3.75-3.75& =& 13.10-3.75\\ 0.85y& =& 9.35\\ \frac{0.85y}{0.85}& =& \frac{9.35}{0.85}\\ y& =& 11\end{array}$

Step 5. Solve for x

Substitute $11$ for $y$ in the second equation and solve for $x$

$$\begin{gathered} x=3y+15 \\ x=3\times 11+15 \\ x=33+15 \\ x=48 \end{gathered}$$

So, he has $48$ quarters and $11$ dimes.

Step 6. Check the solution

Substitute $48$ for $x$ and $11$ for $y$ in the first equation formed.

$\begin{array}{rcl}0.25x+0.10y& =& 13.10\\ 0.25·48+0.10·11& =& 13.10\\ 12+1.10& =& 13.10\\ 13.10& =& 13.10\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}x& =& 3y+15\\ 48& =& 3·11+15\\ 48& =& 33+15\\ 48& =& 48\end{array}$

This is also a true statement.

So the point satisfies both the equations.