Ans: A
$\begin{array}{rcl}
h & = & 3 – \dfrac{5}{k+4} \\
h(k+4) & = & 3(k+4) – 5 \\
hk + 4h & = & 3k + 12 – 5 \\
hk – 3k & = & 7 – 4h \\
k(h – 3) & = & 7 – 4h \\
k & = & \dfrac{7 – 4h}{h – 3} \\
k & = & \dfrac{4h – 7}{ 3 – h}
\end{array}$
$\begin{array}{rcl}
h & = & 3 – \dfrac{5}{k+4} \\
h(k+4) & = & 3(k+4) – 5 \\
hk + 4h & = & 3k + 12 – 5 \\
hk – 3k & = & 7 – 4h \\
k(h – 3) & = & 7 – 4h \\
k & = & \dfrac{7 – 4h}{h – 3} \\
k & = & \dfrac{4h – 7}{ 3 – h}
\end{array}$
