Q. 136 - Intermediate Algebra

Question

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

A cashier has $30$ bills, all of which are \($10$ or \)$20$ bills. The total value of the money is $$460$. How many of each type of bill does the cashier have?

Step-by-Step Solution

Answer

The number of $$10$ bills are $14$ and the number of $$20$ bills are $16$.

Step 1. Given Information

There are total $30$ bills of $$10$ and $$20$.

The total value of the money is $$460$.

Step 2. Identify and name what we are looking for

We need to find the number of $$10$ and $$20$ bills.

Let $x$ represents the number of $$10$ bills

and $y$ represents the number of $$20$ bills.

Step 3. Form the equations

The total number of bills is $30$, so the equation is

$x+y=30 ...\left(1\right)$

Now, the total amount is $$460$ where $x$ is the number of $$10$ bills and $y$ is the number of $$20$ bills. So the equation is

$10x+20y=460 ...\left(2\right)$

Step 4. Solve using substitution

Solve the first equation for  $y$

$\begin{array}{rcl}x+y& =& 30\\ x+y-x& =& 30-x\\ y& =& 30-x ...\left(3\right)\end{array}$

So using the third equation substitute $30-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}10x+20y& =& 460\\ 10x+20(30-x)& =& 460\\ 10x+600-20x& =& 460\\ 10x-20x+600-600& =& 460-600\\ -10x& =& -140\\ \frac{-10x}{-10}& =& \frac{-140}{-10}\\ x& =& 14\end{array}$

Step 5. Find the value of y

Substitute $14$ for $x$ in the third equation

$$\begin{gathered} y=30-x \\ y=30-14 \\ y=16 \end{gathered}$$

So the number of $$10$ bills is $14$ and the number of $$20$ bills is $16$.

Step 6. Check the solution

Substitute $14$ for $x$ and $16$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 30\\ 14+16& =& 30\\ 30& =& 30\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}10x+20y& =& 460\\ 10·14+20·16& =& 460\\ 140+320& =& 460\\ 460& =& 460\end{array}$

This is also a true statement.

So the point satisfies both the equations.