a) ${y^2} + 2x = 13$, $2y + x = 8$

$ \to {y^2} - 4y + 3 = 0$, ${x^2} - 8x + 12 = 0$

$ \to \left( {2,{\text{ }}3} \right)$ and $\left( {6,{\text{ }}1} \right)$

b) Removes $x \to {y^2} + 2\left( {k - 2y} \right) = 13$

Uses ${b^2} - 4\alpha c$ on “quadratic $ = 0$

$ \to k = 8{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$

or $\frac{{dy}}{{dx}} =  - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}} = \frac{{ - 1}}{y} \to y = 2$, $x = 4{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$, $k = 8{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$