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.

Givens 


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*$x\neq 0$


*

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



Goal


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*$y=3$


Proof sketch:


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*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 $x\neq 0$)
