<s>[INST] 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.
 [/INST] $y = \frac{3x^2+2y}{x^2+2}$, 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 $y=3$
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