STORM2K

I have this tough HW problem...can someone please give me a hint on how to solve it?? Here it is:

Tickets for a play cost $6 each for adults and $3 each for children. If
160 of these tickets were bought for a total of $816, how many adult tickets
were bought?

Thanks in advance.

160 x $6 = $960
$960 - $816 = $144
$144 / 3 = 48
160 - 48 = 112
112 x $6 = 672
48 x $3 = 144
672 + 144 = 816

Woohoo! Math!

I love math.

Jassi, what you would do is set up two equations:

Let A be the number of adult tickets and let C be the number of child tickets.

Based on the info you gave,
A + C = 160
and
6A + 3C = $816

Take the top equation and isolate the variable C, and you get
C = 160 - A

Plug that equation into the second equation
(6A + 3C = 816), so you get

6A + 3(160 - A) = 816

From there, solve for A. I got 112.

Hope this helps!

Thanks OtherHD! I can now finally watch the Superbowl without having to worry about HW! That had to be one of the toughest problems I've had this year..and my teacher said they get harder evrey week.

Well I got the answer first and I never took algebra. I just couldn't explain it with letters like OtherHD. BTW are you near the Frederick area in central MD?

Beer980- I'm about 30-40 minutes away from Frederick.

So you might be in the Monrovia area.