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?

Short answer

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

Step-by-step solution

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.