Solving for $y$ in $y= 14x + 1000 = y= 16x + 800$ 
Given $y= 14x + 1000 = y= 16x + 800$, solve for $y$. 

I think I have it: 
$x= 100$. Subbing that in would leave me with:
$y = 14x + 1000$
$y = 14 \cdot 100 + 1000$
$y = 2400$
$y = 16x + 800$
$y = 16 \cdot 100 + 800$
$y = 2400$
the lines break even at $y= 2400$ and $x= 100$
Correct me if i am wrong
 A: $y= 14x + 1000 = y= 16x + 800$
$\implies 14x+1000=16x+800$
Now solve for $x$.
Once you have $x$, you can just plug it back into either equation for $y$. Does that make sense?
EDIT:
You are at this point:
$14x+1000=16x+800$
By solve for $x$, I mean isolate it on one side of the equation. You can do this by eliminating one of the $x$ terms from one side. Let's eliminate $14x$:
$\hspace{3mm}14x+1000=16x+800$
$-14x \hspace{15mm} -14x$
$=0 \hspace{3 mm}+1000=2x+800$ 
$\implies 1000=2x+800$.
Now how do you eliminate the $800$? Let me know if you need more help. 
A: Hint: You know that $y=14x+1000$ and you know that $y=16x+800.$ You want to either use elimination or substitution to solve for either $x$ or for $y$, and then back substitute to find the other variable.
An example is for two equations: $y=5x+2$ and $y=2x-1$. I'll proceed by the substitution method. Since we know both equations are equal to $y$, instead of writing $y$, we can write the other equation instead, since their values are equal to $y$. So we get:
$5x+2=2x-1\\\implies 3x=-3\\\implies x=-1$.
So now we know the $x$ value. We can substitute this into the equation (either equation). I'm going to choose to substitute $x=-1$ into the equation $y=2x-1$. This gives us $y=2(-1)-1=-2-1=-3$. So in my example, $x=-1$ and $y=-3$. The process for your question is similar.
