Question

Need help with Systems of 2 Equations Word Problems . Problem 3.

Answer 1

We are told that a theater have sold 6 adult tickets and 3 child tickets for a total of 96. If "x" is the price for adult and "y" the price for children, then we can describe the situation mathematically as:

$6x+3y=96,(1)$

Now we are told that there were sold 8 adult tickets and 2 children tickets for a total of 112. This can be represented mathematically as:

$8x+2y=112,(2)$

This gives us a system of two equation with two variables. To solve the system we will use the method of subtitution. First, we will solve for "y" in equation (1):

$6x+3y=96$

Subtracting 6x to both sides:

$3y=96-6x$

Dividing both sides by 3:

$\begin{gathered} y=\frac{96}{3}-\frac{6}{3}x \\ y=32-2x \end{gathered}$

Then we will replace this value of "y" in equation (2):

$8x+2(32-2x)=112$

Using the distributive property:

$8x+64-4x=112$

Adding like terms:

$4x+64=112$

Subtracting 64 to both sides:

$4x=112-64$

Solving the operation:

$4x=48$

Dividing both sides by 4:

$x=\frac{48}{4}=12$

Therefore, x = 12. Replacing this value in equation (1):

$y=32-2(12)$

Solving the operations:

$y=8$

Therefore, adult tickets cost $12 and children tickets cost $8.