Algebra Sentence problem part b 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?
 A: Hint: Solve the simultaneous equations
$$f + s = 170$$
$$50 f + 35s = 6625$$
A: 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.
Substitution
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 
$$x + 125 = 170$$ 
$$x = 45$$
Elimination 
$$x + y = 170$$ 
$$50x + 35y = 6625$$ 
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. 
