## Precalculus (6th Edition) Blitzer

The second equation is multiplied by $3$.
Let us consider the provided system: \begin{align} & 3{{x}^{2}}+2{{y}^{2}}=35 \\ & 4{{x}^{2}}+3{{y}^{2}}=48 \end{align} And to eliminate ${{x}^{2}}$, multiply the first equation by $-4$ and the second equation by $3$ and add them to obtain, \begin{align} & -12{{x}^{2}}-8{{y}^{2}}=-140 \\ & \underline{12{{x}^{2}}+9{{y}^{2}}=144} \\ & {{y}^{2}}=4 \\ \end{align} It gives the following values of y: $y=2,-2$ Thus, ${{x}^{2}}$ is eliminated on being multiplied by $3$ in the second equation.