(Without the fancy notation), you are told that the relationship is to be linear, so start with \(\displaystyle y=ax+b\).

Inverting, (solving for x in terms of y) gets you

\(\displaystyle x=\frac{1}{a}y-\frac{b}{a}.\)

and switching the variables

\(\displaystyle y=\frac{1}{a}x-\frac{b}{a}.\)

which is the inverse relationship.

For this to be the same as the original, we need \(\displaystyle a\) to be the same as \(\displaystyle 1/a\) implying that \(\displaystyle a^{2}=1, \text{ so that } a=\pm 1\), and \(\displaystyle b=-b/a\) implying that \(\displaystyle a=-1,\) with \(\displaystyle b\) being able to take any value.

That gets you \(\displaystyle y=b-x\), (from which \(\displaystyle x=b-y).\)