#### Answer

The second equation is multiplied by $3$.

#### Work Step by Step

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.