\( \( \begin{align} {y^2} - y - 3 &= \frac{9}{{{y^2} - y - 3}}\\ {\rm{Stel }}{y^2} - y - 3 &= k\\ k &= \frac{9}{k}\\ KGV &= k\\ BPK:k &\ne 0\\ {k^2} &= 9\\ {k^2} - 9 &= 0\\ \left( {k - 3} \right)\left( {k + 3} \right) &= 0\\ k &= 3&k &= - 3\\ {\rm{maar }}\,\,k &= {y^2} - y - 3\\ \therefore 3 &= {y^2} - y - 3 & - 3 &= {y^2} - y - 3\\ 0 &= {y^2} - y - 6 & 0 &= {y^2} - y\\ 0 &= \left({y - 3}\right)\left({y + 2}\right) & 0&= y\left( {y - 1} \right)\\ \therefore y &= 3\,\,{\rm{ of }}\,\,y = - 2 & \therefore y &= 0\,\,{\rm{ of }}\,\,y = 1 \end{align} \) \)