Answer
The points $\left( 2,1 \right),\left( 2,-1 \right),\left( -2,1 \right)$ and $\left( -2,-1 \right)$ are the solutions of both equations.
Work Step by Step
The given set of equations is\[\begin{align}
& 2{{x}^{2}}-3{{y}^{2}}=5 \\
& 3{{x}^{2}}+4{{y}^{2}}=16
\end{align}\].
Calculation:
The provided equations are,
$3{{x}^{2}}+4{{y}^{2}}=16$ (I)
$2{{x}^{2}}-3{{y}^{2}}=5$ (II)
Simplify as follows:
Step 1:
Let’s eleminate the variable ${{y}^{2}}$, multiply 3 in equation (I) and multiply 4 in equation (2), by addition method in equation (I) and (II) simplify as follows:
\[\begin{align}
& 9{{x}^{2}}+12{{y}^{2}}+\left( 8{{x}^{2}}-12{{y}^{2}} \right)=48+20 \\
& 17{{x}^{2}}=68 \\
& {{x}^{2}}=4
\end{align}\]
So, $x=\pm 2$ are solutions of both equations.
Step 2:
Put the value of $x$ in equation (I) to find out the value of $y$ simplify as follows:
When$x=\pm 2$,
$\begin{align}
& 3{{x}^{2}}+4{{y}^{2}}=16 \\
& 3{{\left( \pm 2 \right)}^{2}}+4{{y}^{2}}=16 \\
& 12+4{{y}^{2}}=16 \\
& 4{{y}^{2}}=4
\end{align}$
Simplify further,
${{y}^{2}}=1$
So, $y=\pm 1$ are solutions of this equations.
Step 3:
Check the proposed solutions in both of the equations, now begin by check\[\left( \pm 2,\pm 1 \right)\], and replace $x$ with \[\pm 2\] and $y$ with\[\pm 1\].
\[\begin{align}
& 3{{x}^{2}}+4{{y}^{2}}=16 \\
& 3{{\left( \pm 2 \right)}^{2}}+4{{\left( \pm 1 \right)}^{2}}=16 \\
& 12+4=16 \\
& 16=16
\end{align}\]
$\text{True}$
And,
\[\begin{align}
& 2{{x}^{2}}-3{{y}^{2}}=5 \\
& 2\times {{\left( \pm 2 \right)}^{2}}-3\times {{\left( \pm 1 \right)}^{2}}=5 \\
& 8-3=5 \\
& 5=5
\end{align}\]
$\text{True}$
Hence, the points $\left( 2,1 \right),\left( 2,-1 \right),\left( -2,1 \right)$ and $\left( -2,-1 \right)$ are the solution of the system.
