Q 4.49 - Intermediate Algebra

Question

Translate to a system of equations and solve:

Matilda has a handful of quarters and dimes, with a total value of $\$8.55$. The number of quarters is $3$ more than

twice the number of dimes. How many dimes and how many quarters does she have?

Step-by-Step Solution

Answer

The number of quarters$=29$

The number of dimes$=13$

Step 1. Given information

• The total worth of quarters and dimes is $\$8.55$.
• It is given that the number of quarters are $3$ more than twice the number of dimes.

Step 2. Form the equations

We know that the value of each quarter is $\$0.25$ and the value of each dime is $\$0.10$,

Let the number of quarters be $q$ and the number of dimes be $d$.

According to the question, we get:

$0.25q+0.10d=8.55\_\_\_\left(1\right)$ .

Now, it is given that the number of quarters are $3$ more than twice the number of dimes,

So,

$q=2d+3\_\_\_\left(2\right)$

Step 3. Solve the equations by substitution method

Substituting the equation $\left(2\right)$ in equation $\left(1\right)$, we get:

$$\begin{gathered} 0.25q+0.10d=8.55 \\ 0.25(2d+3)+0.10d=8.55 \\ 0.50d+0.75+0.10=8.55 \\ 0.60d=8.55-0.75 \\ 0.6d=0.78 \end{gathered}$$

Dividing $0.60$ into both sides of the equation, we get:

$$\begin{gathered} \frac{0.60d}{0.60}=\frac{7.8}{0.60} \\ d=13 \end{gathered}$$

Step 4. Find the value of q

Substituting the value of $d=13$ in the equation $\left(2\right)$, we get:

$$\begin{gathered} q=2d+3 \\ q=2\times 13+3 \\ q=26+3 \\ q=29 \end{gathered}$$

So, Matlida has $29$ quarters and $13$ dimes.