Prove that there is a unique real number $x$ such that for every real number $y$, $xy + x -17 = 17y$. My attempt is as follows.
I factored $xy + x - 17 - 17y$ into $(x-17)(y+1)$ and got solutions: $x=17,y=-1$.
So in this case; Is my unique value for $x$, $x=17$.
I tried using $y=-1$ but it just reduced $xy + x - 17 = 17y$ to $0=0$.
Am I approaching this problem correctly?
 A: You have the germ of a correct idea, but as written it's not at all correct. You say

I factored $xy + x - 17 = 17 y$ into $(x - 17)(y + 1)$ and got solutions...

where what you actually mean is that you rearranged the equation as
$$xy + x - 17 - 17 y = 0$$
and then factored. Then you used $y = -1$, but you need to prove that something is true for all $y$ - that is, you don't get to select $y$; furthermore, reducing the equation to $0 = 0$ doesn't tell you much. Remember that the statement you're trying to prove is that there exists $x$ such that for all $y$ something happens. You're supposed to say what that $x$ is.

But you do have part of the correct idea. You found something special about $x =17$, so your work from here should proceed along the lines:
Let $x = 17$, and let $y$ be arbitrary. Then 
$$xy + x - 17 = 17 y + 17 - 17 = 17 y$$
as desired.
You also have the uniqueness issue, and for this one you can use what happens when $y \ne -1$; you already know this forces $x = 17$.
A: Your expression is
$$(x-17)y+x-17=0$$
it doesn't depend on $y $ if the coefficient of $y $ is zero.
this gives $$x-17=0$$
and $x=17$.
A: First let's translate the problem into its logical form (if you want to do it more formally) which will be: $∃!x∀y(xy + x - 17 = 17y)$. One short note: $∃!x$ means that "there exists a unique $x$ such that...".
Proof. 
Existence. Let $x$ be 17. Then $xy + x -17 = 17y + 17 - 17 = 17y$. Ergo there exists an $x$ such that the above equation holds.
Uniqueness. Let $z$ be a real number such that $zy + z -17 = 17y$. Then $zy + z = 17 + 17y$ and $z(y + 1) = 17(y + 1)$. Now we have two cases:
Case 1. $y = -1$, then any value of $z$ works.
Case 2. $y \neq -1$, then we can conclude that $z = 17 = x$.
Therefore there exists a unique value of $x$ such that our equality holds for all value of y.
A: We have $$(x-17)(y+1)=0.$$
If $x=17$ we obtain
$$0(y+1)=0,$$
which is true for all real $y$,
which says that $x=17$ is valid.
In another hand, if $x\neq17$ we have
$$y=-1,$$  which is not true for all value of $y$.
Thus, $17$ it's an unique value of $x$, for which our equality is true for all value of $y$.
Done!
