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Here is a new one I need help with.

A dinner theater sells two types of tickets. Floor seats cost \$50 while stadium seats cost \$35. If the theater sold 170 seats for \$6625, how many tickets of each type were sold?

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$$x + y = 170$$ $$50x + 35y = 6625$$ can you solve this system of equations? – Deven Ware Sep 21 '11 at 0:57
No , i need help, please. – naknode Sep 21 '11 at 1:38
up vote 4 down vote accepted

Hint: Solve the simultaneous equations

$$f + s = 170$$ $$50 f + 35s = 6625$$

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But... assuming s is 0... you can't divide 50 into 6625 evenly... – naknode Sep 21 '11 at 1:17
and the same thing goes for f... – naknode Sep 21 '11 at 1:18
Why would we assume that $s = 0$ ?? From the first equation, if $s = 0$, then $f = 170$. But these values of s, f do not make the second equation true. Please research (on Youtube or the topic "solving systems of equations" – The Chaz 2.0 Sep 21 '11 at 2:02

I can see from your reply to my comment that the problem is not writing down the system of equations based on the word problem, but rather solving the system.

You have already acknowledge you understand that we have:

$$x + y = 170$$

$$50x + 35y = 6625$$

Now, for two variable systems of equations there are usually two approaches I'd take a look at elimination or substitution. The most useful is usually elimination but I will show a solution via both methods here.


We can rearrange the first equation and see that $x = 170 - y$. Now we should substitute this in to the second equation in order to solve the system.

$$50(170-y) + 35y = 6625$$

$$ 8500 - 50y + 35y = 6625$$

$$8500 - 6625 - 15y = 0$$

$$1875 = 15y$$

$$125 = y$$

Then we can plug this solution for $y$ back into the first equation and see

$$x + 125 = 170$$

$$x = 45$$


$$x + y = 170$$ $$50x + 35y = 6625$$

When solving by elimination we should look to multiply one equation by a number such that if we add it to the second equation, only one variable remains. i.e we should look to multiply one equation by a number in such a way that when we add the two equations, we eliminate a variable. In this situation we could multiply the first equation by $(-35)$ and we will be able to eliminate $y$.

$$(-35x - 35y = -5950)$$

$$+ (50x + 35y = 6625)$$

$$ 15x = 675$$

$$x = 45$$

Then we can plug this back into the first equation

$$45 + y = 170$$

$$y = 125$$

Which is the same solution we got from the alternate method of substitution.

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