For all real numbers x and y there is a real number $z$ such that $x + z = y − z$.

Question

To Prove: For all real numbers $x$ and $y$ there is a real number $z$ such that $x + z = y − z$.

Proof:

$x+z=y-z \Rightarrow y-x=2z$. Since $y$ and $x$ are real numbers, $2z$ is also real. Therefore $z$ is real.

Is that good enough?

Answer 1

The proof you have written is really the "scratch work" you should do before writing the proof down. In particular, the proof itself should not begin with the assumption that $x+z=y-z$. (What is $z$? It hasn't been defined. How do you know such $z$ exists? That's what you're trying to prove, after all.)

So your "scratch work" should look like:

If $x+z=y-z$, then $2z=y-x$, so $z=\frac{1}{2}(y-x)$.

And your proof should look like:

Proof: Let $x$ and $y$ be real numbers. Let $z=\frac{1}{2}(y-x)$, and note that $z$ is a real number. Then... [here you should verify that $x+z=y-z$].

Answer 2

You only found a $z$ which might work to fulfill the equality. But you need to verify that this $z$ works indeed.

So you have $z=\frac{y-x}{2}$ and $x+z=\frac{y-x+2x}{2}=\frac{y+x}{2}=\frac{2y+x-y}{2}=y-z$, so your $z$ works, so you proved that there exists such a $z$.