Q. 128 - Intermediate Algebra

Question

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

Tickets for an Amtrak train cost \($10$ for children and \)$22$ for adults. Josie paid $$1200$ for a total of $72$ tickets. How many children tickets and how many adult tickets did Josie buy? 

Step-by-Step Solution

Answer

Josie bought $32$ tickets for children and $40$ tickets for adults.

Step 1. Given Information

Given that the price of one adult ticket is $$22$ and the price of one child ticket is $ $10$.

The total number of tickets sold is $72$ and the amount collected is $$1200$

Step 2. Identify and name what we are looking for

The objective is to find the number of adult and child tickets sold.

Let the number of adult tickets sold be $x$ and the number of child tickets sold be $y$

Step 3. Form the equations

The total number of tickets sold is $72$. So an equation can be formed as

$x+y=72 ...\left(1\right)$

Now the total cost of $x$ adult tickets each at $$22$ and $y$ child tickets each at $$10$ is $$1200$. So another equation can be written as

$22x+10y=1200 ...\left(2\right)$

Step 4. Solve using substitution

The first equation can be solved for $y$ as

$\begin{array}{rcl}x+y& =& 72\\ x+y-x& =& 72-x\\ y& =& 72-x ...\left(3\right)\end{array}$

So using the third equation substitute $72-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}22x+10y& =& 1200\\ 22x+10(72-x)& =& 1200\\ 22x+720-10x& =& 1200\\ 22x-10x+720-720& =& 1200-720\\ 12x& =& 480\\ \frac{12x}{12}& =& \frac{480}{12}\\ x& =& 40\end{array}$

Step 5. Find the value of y

Substitute $40$ for $x$ in the third equation and find the value of $y$

$$\begin{gathered} y=72-x \\ y=72-40 \\ y=32 \end{gathered}$$

So $40$ tickets for adults and $32$ tickets for children were bought.

Step 6. Check the solution

Substitute $40$ for $x$ and $32$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 72\\ 40+32& =& 72\\ 72& =& 72\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}22x+10y& =& 1200\\ 22·40+10·32& =& 1200\\ 880+320& =& 1200\\ 1200& =& 1200\end{array}$

This is also a true statement.

So the point satisfies both the equations.