Answer
True statement
Work Step by Step
Let the two second-degree equations are
$3{{x}^{2}}+2{{y}^{2}}=12$ …… (1)
${{x}^{2}}+2{{y}^{2}}=4$ …… (2)
Subtract the equations:
$\begin{align}
& \left( 3{{x}^{2}}-{{x}^{2}} \right)+\left( 2{{y}^{2}}-2{{y}^{2}} \right)=12-4 \\
& 2{{x}^{2}}+0=8 \\
& {{x}^{2}}=\frac{8}{2} \\
& {{x}^{2}}=4
\end{align}$
Simplify more,
$\begin{align}
& {{x}^{2}}=4 \\
& x=\pm \sqrt{4} \\
& x=\pm 2
\end{align}$
$x=2$
And
$x=-2$
Substitute $x=2$ in equation (2), and solve for y:
$\begin{align}
& {{x}^{2}}+2{{y}^{2}}=4 \\
& {{\left( 2 \right)}^{2}}+2{{y}^{2}}=4 \\
& 4+2{{y}^{2}}=4 \\
& 2{{y}^{2}}=0
\end{align}$
$\begin{align}
& 2{{y}^{2}}=0 \\
& {{y}^{2}}=0 \\
& y=0
\end{align}$
The ordered pair is $\left( 2,0 \right)$.
Substitute $x=-2$ in equation (1), and solve for y:
$\begin{align}
& 3{{x}^{2}}+2{{y}^{2}}=12 \\
& 3{{\left( -2 \right)}^{2}}+2{{y}^{2}}=12 \\
& 12+2{{y}^{2}}=12 \\
& 2{{y}^{2}}=0
\end{align}$
$\begin{align}
& 2{{y}^{2}}=0 \\
& {{y}^{2}}=0 \\
& y=0
\end{align}$
The ordered pair is $\left( -2,0 \right)$.
Thus, the elimination method is best.
