91Ó°ÊÓ

We have given the following system of equations :-

$\left\{\begin{array}{l}{x}^2+y^2=25\\ 2x^2-3y^2=5\end{array}\right.$

We have to solve this system of equations. That is we have to find pair for values of $xandy$that satisfies this system of equations.

We will apply some arithmetic operation to eliminate one variable. Then we solve the equation for remaining variable.

After that we put the values in the any of the given equation to get the values of first variable as well.

The given system of equations is :-

$\left\{\begin{array}{l}{x}^2+y^2=25\\ 2x^2-3y^2=5\end{array}\right.$

Multiply the first equation with 2, then we have :-

$\left\{\begin{array}{l}2x^2+2y^2=50\\ 2x^2-3y^2=5\end{array}\right.$

Now, the coefficient of $x^2$are of same sign. So we will subtract the second equation from first, then we have :-

$$\begin{gathered} 2x^2+2y^2-(2x^2-3y^2)=50-5 \\ ⇒2x^2+2y^2-2x^2+3y^2=45 \\ ⇒5y^2=45 \\ ⇒y^2=9 \\ ⇒y=Â\pm \sqrt{9} \\ ⇒y=Â\pm 3 \end{gathered}$$

We find :-
$y=Â\pm 3$

Also we have :-

$x^2+y^2=25$

So that when $y=Â\pm 3$,

$$\begin{gathered} x^2+{\left(Â\pm 3\right)}^2=25 \\ ⇒x^2+9=25 \\ ⇒x^2=25-9 \\ ⇒x^2=16 \\ ⇒x=Â\pm \sqrt{16} \\ ⇒x=Â\pm 4 \end{gathered}$$

Then we have following four solution pairs :-

$\left(4,3\right),\left(-4,3\right),\left(4,-3\right)and\left(-4,-3\right)$.