## Precalculus (6th Edition) Blitzer

The solution of the system is $\underline{\left\{ \left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right) \right\}}$
Let us consider the system of the given equations: $2{{x}^{2}}-5{{y}^{2}}=-2$ (I) $3{{x}^{2}}+2{{y}^{2}}=35$ (II) Rewrite equation (I) as given below: \begin{align} & 2{{x}^{2}}-5{{y}^{2}}=-2 \\ & 2{{x}^{2}}=5{{y}^{2}}-2 \\ & {{x}^{2}}=\frac{5{{y}^{2}}-2}{2} \end{align} Put the value of x in equation (II) to obtain the value of y as follows: \begin{align} & 3\left( \frac{5{{y}^{2}}-2}{2} \right)+2{{y}^{2}}=35 \\ & 15{{y}^{2}}-6+4{{y}^{2}}=70 \\ & 19{{y}^{2}}=76 \\ & {{y}^{2}}=4 \end{align} $y=\pm 2$ Put the value of ${{y}^{2}}$ in equation (II) to obtain the value of x as follows: \begin{align} & 3{{x}^{2}}+2\left( 4 \right)=35 \\ & 3{{x}^{2}}+8=35 \\ & 3{{x}^{2}}=27 \\ & {{x}^{2}}=9 \end{align} $x=\pm 3$ Thus, the solutions of the equations are $\left( x,y \right)=\left\{ \left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right) \right\}$.