91Ó°ÊÓ

We have given the following system of equations:-

$\left\{\begin{array}{l}{x}^2+y^2=16\\ x^2-y=4\end{array}\right.$

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 operations to eliminate one variable. Then we solve the equation for the remaining variable.

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

The given system of equations is :-

$\left\{\begin{array}{l}{x}^2+y^2=16\\ x^2-y=4\end{array}\right.$

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

$$\begin{gathered} x^2+y^2-(x^2-y)=16-4 \\ {x}^2+y^2-x^2+y=12 \\ {y}^2+y-12=0 \end{gathered}$$

We have the following quadratic equation :-

$$\begin{gathered} y^2+y-12=0 \\ ⇒y^2+4y-3y-12=0 \\ ⇒y(y+4)-3(y+4)=0 \\ ⇒(y+4)(y-3)=0 \\ ⇒y+4=0ory-3=0 \\ ⇒y=-4ory=3 \\ ⇒y=-4,3 \end{gathered}$$

We find :-

$y=-4,3$

Also we have :-

$x^2+y^2=16$.

So that :-

When role="math" localid="1645366305500" $y=-4$,

$$\begin{gathered} x^2+{(-4)}^2=16⇒x^2+16=16 \\ ⇒x^2=16-16 \\ ⇒x^2=0 \\ ⇒x=0 \end{gathered}$$

So the solution pair is $\left(0,-4\right)$.

When role="math" localid="1645366296477" $y=3$,

$$\begin{gathered} x^2+3^2=16 \\ {x}^2+9=16 \\ {x}^2=16-9 \\ {x}^2=7 \\ x=Â\pm \sqrt{7} \end{gathered}$$

So the solution pairs are$\left(\sqrt{7},3\right)and\left(-\sqrt{7},3\right)$.