$$x = \frac{{6 - 10y}}{{8y + 7}}$$ Show Step 3

$$\begin{align*}x & = \frac{{6 - 10y}}{{8y + 7}}\\ x\left( {8y + 7} \right) & = 6 - 10y\\ & 8xy + 7x = 6 - 10y\\ 8xy + 10y & = 6 - 7x\\ & \left( {8x + 10} \right)y = 6 - 7x\\ y & = \frac{{6 - 7x}}{{8x + 10}}\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{f^{ - 1}}\left( x \right) = \frac{{6 - 7x}}{{8x + 10}}}}\end{align*}$$

Note that the Algebra in these kinds of problems can often be fairly messy, but don’t let that make you decide that you can’t do these problems. Messy Algebra will be a fairly common occurrence in a Calculus class so you’ll need to get used to it!

Finally, compute either $\left( {f \circ {f^{ - 1}}} \right)\left( x \right)$ or $\left( {{f^{ - 1}} \circ f} \right)\left( x \right)$ to verify our work.

Show Step 4

Either composition can be done so let’s do $\left( {f \circ {f^{ - 1}}} \right)\left( x \right)$ in this case. As with the previous step, the Algebra here is going to be messy and in fact will probably be messier.

$$\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) & = f\left[ {{f^{ - 1}}\left( x \right)} \right]\\ & = \frac{{6 - 10\left[ {\frac{{6 - 7x}}{{8x + 10}}} \right]}}{{8\left[ {\frac{{6 - 7x}}{{8x + 10}}} \right] + 7}}\,\,\,\frac{{8x + 10}}{{8x + 10}}\\ & = \frac{{6\left( {8x + 10} \right) - 10\left( {6 - 7x} \right)}}{{8\left( {6 - 7x} \right) + 7\left( {8x + 10} \right)}}\\ & = \frac{{48x + 60 - 60 + 70x}}{{48 - 56x + 56x + 70}}\\ & = \frac{{118x}}{{118}}\\ & = x\end{align*}$$

So, we got $x$ out of the composition and so we know we’ve done our work correctly.