How do I prove that for any $y$ $\in$ $B$, $y$ $\in$ $f$($A$)?

Question

In order to show (c), I know that I need to show that for every element, y $\in$ $B$, y $\in$ $f$($A$). I take $y$ to be arbitrary and show $y$ $\in$ $f$($A$) by showing it is in the universal set and that it satisfies the given condition. So, I pick some $y$ $\in$ $B$, namely, y = $\frac{2x-1}{3x+2}$ and solve for $x$ to get $x$ = $\frac{2y+1}{3y+2}$. I am unsure of how to proceed from here to show that y $\in$ $f$($A$) and that it satisfies the condition.

Answer 1

You get $x=\dfrac{2y+1}{2-3y}$. Just plug this into $f(x)$ to show the output is $y$ which shows $y\in f(A)$. Also observe $y\neq\dfrac23$ as $x$ will be undefined. Hence $y\in A$.

Answer 2

You have made a mistake in calculating $x$. You should get $x=\frac {2y+1} {2-3y}$. Given $y \neq \frac 2 3$ you have to show that $x$ obtained from $x=\frac {2y+1} {2-3y}$ belongs to $A$ which means $x \neq -\frac 2 3$. Prove this by contradiction. Suppose $x=-\frac 2 3$. Then $-2(2-3y)=3(2y+1)$ which simplfies to $-4=3$ which is a contradiction.