## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.
Let us consider the given equation: $3{{x}^{2}}+4{{y}^{2}}=16$ …… (I) $2{{x}^{2}}-3{{y}^{2}}=5$ …… (II) By multiplying equation (I) by $3$, equation (II) by $4$ and add equation (I) and equation (II): we get, \begin{align} & 9{{x}^{2}}+12{{y}^{2}}=48 \\ & \underline{4{{x}^{2}}-12{{y}^{2}}=20} \\ & 17{{x}^{2}}=64 \\ & {{x}^{2}}=4 \\ & x=\pm 2 \\ \end{align} Substitute $x=2$ in equation (I): \begin{align} & 3{{\left( 2 \right)}^{2}}+4{{y}^{2}}=16 \\ & {{y}^{2}}=2 \\ & y=\pm 1 \end{align} Substitute $x=-2$ in equation (II): \begin{align} & 3{{\left( -2 \right)}^{2}}+4{{y}^{2}}=16 \\ & {{y}^{2}}=2 \\ & y=\pm 1 \end{align} Thus, the solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.