## Precalculus (6th Edition) Blitzer

The values of x, y are $\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$
$4{{x}^{2}}+3{{y}^{2}}=48$ (I) $3{{x}^{2}}+2{{y}^{2}}=35$ (II) And subtract $3$ times (I) from $4$ times (II), to get: \begin{align} & 12{{x}^{2}}+8{{y}^{2}}-12{{x}^{2}}-9{{y}^{2}}=140-144 \\ & -{{y}^{2}}=-4 \\ & y=\sqrt{4} \\ & y=\pm 2 \end{align} To get ${{x}^{2}}$ substitute the value of ${{y}^{2}}=4$ in (I): \begin{align} & 4{{x}^{2}}+3\left( 4 \right)=48 \\ & 4{{x}^{2}}=48-12 \\ & {{x}^{2}}=\frac{36}{4} \\ & {{x}^{2}}=9 \end{align} Therefore, $x=\pm 3$ So, the pairs of $\left( x,y \right)$ are as follows: $\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$ Thus, the values of x, y are $\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$.