If f(x) = (3x-5)/(2), what is f^-1(7)?

Set $y=\frac{3x-5}{2}$ and swap the variables: $x=\frac{3y-5}{2}$. Then $2x=3y-5$, so $y=\frac{2x+5}{3}$. At $x=7$: $f^{-1}(7)=\frac{14+5}{3}=\frac{19}{3}$, choice B. Check: $f\left(\frac{19}{3}\right)=\frac{19-5}{2}=7$. Choice A ($3$) subtracts $5$ instead of adding it. Choice C ($\frac{29}{3}$) adds $15$ instead of $5$. Choice D ($9.5$) divides by $2$ instead of $3$. Each wrong choice comes from an error while undoing the operations in the right order, so always verify by plugging your answer back into $f$. The final check takes only one line, so it is always worth doing.