Monday

November 24, 2014

November 24, 2014

Posted by **ANYONE Smart** on Sunday, November 25, 2007 at 8:20pm.

- algebra -
**Michael**, Sunday, November 25, 2007 at 8:25pmSet up a system of equations to model the situation.

Let a be the number of adult tickets sold, and let c be the number of child tickets sold.

See what you get, and I'll help you.

- algebra -
**ANYONE Smart**, Sunday, November 25, 2007 at 8:28pma+c=20??

- algebra -
**Michael**, Sunday, November 25, 2007 at 8:29pmThat's one of your equations, but a system of equations is two equations.

Find the other using the ticket pricing.

- algebra -

- algebra -
- algebra -
**ANYONE Smart**, Sunday, November 25, 2007 at 8:31pm13a+3c=200??

- algebra -
**Michael**, Sunday, November 25, 2007 at 8:34pmYes, very good. Now, we are trying to find the number of children who attended the play (c).

We have these equations:

a + c = 20

13a + 3c = 200

The easiest way to solve this is substitution. Solve the first equation for a, and then plug that into the second equation. You will then have all c's in the equation and will be able to solve it like any other equation.

- algebra -
- algebra -
**kim**, Saturday, October 25, 2014 at 11:33ama=20-c

13(20-c)+3c = 200

260 -13c +3c = 200

60=10c

6=c

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