Question

Mr. Kaplan bought 11 tickets to the circus and spent $50. He bought child tickets for $4 each and bought adult tickets for $7 each. How many child and adult tickets did Mr. Kaplan buy?

Answer 1

9 child and 2 adult

Explanation:

We can set up a systems of equations to solve this.

First equation:

We know he spent a total of $50 and child tickets are $4 while adult tickets are $7. We can set up an equation to represent this.

50=4c + 7a

c represents the number of child tickets bought and a represents the number of adult tickets bought.

Second Equation:

We know he bought 11 tickets, some of them are child tickets while others are for adults. We can represent this in an equation.

11= c + a

c represents the number of child tickets bought and a represents the number of adult tickets bought.

Put it all together:

The first step to solve one of the equations, isolating one of the variables. In this explanation, I will be solving the second equation first. I am going to solve it to be in terms of a, meaning I will isolate a.

1. 11= c + a
2. Subtract c from both sides
3. Now you have 11 - c = a

Now that I have this equation, I will plug it in for a in the first equation and solve to isolate c.

1. 50 = 4c + 7 (11 - c)
2. Distribute the 7: 50= 4c + 77 - 7c
3. Now combine like terms: 50 = -3c + 77
4. Subtract 77 from both sides: -27 = -3c
5. Divide both sides by -3: 9=c

Now we know that the number of child tickets bought was 9. To find out the number of adults tickets bought we plug 9 into another equation. I'm going to plug it into the second equation and solve to isolate a.

1. 11= 9 + a
2. Subtract 9 from both sides: 2= a

Now we are done, we know he bought 2 adult tickets and 9 child tickets.