Q. 132 - Intermediate Algebra

Question

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

Brandon has a cup of quarters and dimes with a total value of $$3.80$. The number of quarters is four less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

Step-by-Step Solution

Answer

Brandon has $12$ quarters and $8$ dimes.

Step 1. Given Information

Given that the total values of quarters and dimes is $$3.80$

Also, the number of quarters is four less than twice the number of dimes.

Step 2. Identify and name what we are looking for

We need to find the number of quarters and dimes Brandon 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 $$3.80$. So an equation can be written as

$0.25x+0.10y=3.80 ...\left(1\right)$

The number of quarters is four less than twice the number of dimes, so it can be written as

$x=2y-4 ...\left(2\right)$

Step 4. Solve using substitution

Using the second equation, substitute $2y-4$ for $x$ in the first equation and solve for $y$

$\begin{array}{rcl}0.25x+0.10y& =& 3.80\\ 0.25(2y-4)+0.10y& =& 3.80\\ 0.50y-1+0.10y& =& 3.80\\ 0.50y+0.10y-1+1& =& 3.80+1\\ 0.60y& =& 4.80\\ \frac{0.60y}{0.60}& =& \frac{4.80}{0.60}\\ y& =& 8\end{array}$

Step 5. Solve for x

Substitute $8$ for $y$ in the first equation and find the value of $x$

$$\begin{gathered} x=2y-4 \\ x=2\times 8-4 \\ x=16-4 \\ x=12 \end{gathered}$$

So Brandon has $12$ quarters and $8$ dimes.

Step 6. Check the solution

Substitute $12$ for $x$ and $8$ for $y$ inthe first equation formed.

$\begin{array}{rcl}0.25x+0.10y& =& 3.80\\ 0.25·12+0.10·8& =& 3.80\\ 3+0.80& =& 3.80\\ 3.80& =& 3.80\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}x& =& 2y-4\\ 12& =& 2·8-4\\ 12& =& 16-4\\ 12& =& 12\end{array}$

This is also a true statement.

So the point satisfies both the equations.