Question

There are 300 people at a play. Some paid $5.00 for reserved seats; others paid $3.00 for general admission seats. Total receipts were $1080. How many tickets of each kind were sold?

Answer 1

r+g=300, r=300-g

5r+3g=1080, using r found above in this you get:

5(300-g)+3g=1080

1500-5g+3g=1080

1500-2g=1080

-2g=-420

g=210, since r=300-g

r=90

So 90 reserved and 310 general tickets were sold.

Answer 2

What we will have to do in this case is probably create a system of equations.
First, assign the variables to what we're trying to find.

Let x = the number of general admission tickets sold
Let y = the number of reserved tickets sold

Then x + y = the total number of tickets sold
Then 3x + 5y = how much money was made

Here is our system of equations:

x + y = 300
3x + 5y = 1080

Let's first solve for x. That means will have to cancel out the y variables in both of the equations when we add them up together.

Multiply the top equation by -5.

-5(x + y) = 300 * -3 = -5x - 5y = -1500

Now we have the system of equations.

-5x - 5y = -1500
3x + 5y = 1080

Add them up together to get one equation.

When you get the sum of the left-hand side, the y's cancel out.

-5x - 5y + 3x + 5y = -2x + 0y or just -2x

Now add up the right-hand side.

-1500 + 1080 = -420

Now we have the equation -2x = -420. Solve for x.

Divide both sides by -2.

-2x / -2 = -420 / -2
x = 210

Now we know how many general admission tickets were sold.
But what about the reserved tickets? Simply replace the x variable with the value of x in one of the equations.

I'll pick x + y = 300 because it's simpler.

Replace x with the value of x.

210 + y = 300

Solve for y.

Subtract both sides by 210.

210 + y - 210 = 300 - 210
y = 90

So, 210 general admission tickets were sold and 90 reserved tickets were sold.