smc

Because $AC=12$ and $AM=x$, $MC=12-x$. Let $MN=z$. Then, because $MNPC$ is a rectangle, $NP=12-x$ and $PC=z$, and so $BP=5-z$. By AA similarity, $△AMN\sim △NPB$. From this similarity, we can solve the following proportion for $z$: $$\begin{array}{rl}\frac{BP}{PN}& =\frac{NM}{AM}\\ \frac{5-z}{12-x}& =\frac{z}{x}\\ 5x-xz& =12z-xz\\ 5x& =12z\\ z& =\frac{5x}{12}\end{array}$$ Because $y=MN+NP=z+12-x$, we can now substitute for $z$ and find $y$ in terms of $x$: $$\begin{array}{rl}y& =z+12-x\\ & =\frac{5x}{12}-x+12\\ & =\frac{-7x}{12}+\frac{144}{12}\\ & =\frac{144-7x}{12}\end{array}$$ Thus, our answer is $\boxed{y=\frac{144-7x}{12}}$.