How can I prove $f(x) = \frac32 x − \frac23$ is surjective? So I'm trying to prove the function $f(x) = \frac32 x − \frac23$ is bijective (injective and surjective)...
For injection I can easily write $\frac32x - \frac23 = \frac32y - \frac23$; that gives me $x=y$ so it's injective.
So the one more step to prove is to show function is surjective. I can say $f(x)=y$ which gives me $\frac32 x - \frac23 = y$ or $\frac{9x - 4}{6} = y$,
which finally gives me $x = \frac{6y+4}{9}$.
But there is a problem when I plug $x$ in function $\frac32 x - \frac23$ I can't get that $y$ value... Basically I get $\frac32 (6y + 4)\frac19 - \frac23 \implies 6y + \frac23 - \frac23$ which is $6y$ but I need $y$? I saw on Wolfram that this function must be surjective but I don't know how to prove it. Can anyone help me with that. And almost forgot to mention $f :\mathbb R \to\mathbb R$.
Thanks in advance :)

The function is surjective, you just made a small computational error.
\begin{align*}
\frac{3}{2} \frac{6y+4}{9} - \frac{2}{3} &= \frac{6y+4}{6} - \frac{2}{3}\\
&= y + \frac{2}{3}-\frac{2}{3}\\
&= y.
\end{align*}
