Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$ The problem is the following (Velleman's exercise 3.2.10): 

Suppose that $x$ and $y$ are real numbers. Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$.

My approach so far: Suppose $x \neq 0$. Suppose $ y = \frac{3x^2+2y}{x^2+2}$. Suppose  $y \neq 3$. Suppose $y=4$. Then $4x^2=3x^2 \iff x=0$. Contradiction to $x \neq 0$.
Is it correct? I am a noob so feel free to bash it as good as you can. Thanks in advance.
 A: $y = \frac{3x^2+2y}{x^2+2}$, multiplying both sides of the equation by $x^2+2$ results in an equivalent equation because that term is never $0$ (in the reals at least).
You end up with $yx^2 + 2y = 3x^2+2y$
subtract $2y$ from both sides (always legitimate).
$yx^2=3x^2$
Since $x\neq 0$ we can divide both sides by $x^2$ and get $y=3$
A: Your proof is incorrect. Under your assumptions, if $y\neq 3$ you want to find a contradiction but you choose a special case: $y=4$. So all you proved is that $y$ can't be $4$, not $y$ must be $3$. So if you want to use a proof by contradiction, then you should work with any $y\neq 3$. But this question doesn't need an indirect proof. Work on $y=\dfrac{3x^2+2y}{x^2+2}$. See it as an equation of unknown $y$ and $x$ a non-zero number. You have $y=\dfrac{3x^2+2y}{x^2+2}\Rightarrow y(x^2+2)=3x^2+2y$. Carry on!
A: Givens 


*

*$x\neq 0$


*

*$y = \frac{3x^2+2y}{x^2+2}$



Goal


*

*$y=3$


Proof sketch:


*

*Suppose $x\neq 0$. 


*Suppose $y = \frac{3x^2+2y}{x^2+2}$. This means that $y (x^2+2)=3x^2+2y \Rightarrow x^2y+2y=3x^2+2y $.



$\qquad \quad$ [Proof of $y=3$ goes here.]
$\qquad$ 3. Therefore $y=3$ 


*Thus, if $x\neq 0$, then if $y = \frac{3x^2+2y}{x^2+2}$, then $y=3$


The key is the algebraic manipulation to get $x^2y+2y=3x^2+2y$. Now can you get rid of the $2y$ and $x^2$ on both sides? (for the latter, recall that $x\neq 0$)
