91Ó°ÊÓ

We have given the following system of equations :-

$\left\{\begin{array}{l}{x}^2+y^2=25\\ 2x+y=10\end{array}\right.$

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 substitute the value of$y$from second equation in first one. Than we have a quadratic equation in $x$. After that we will solve the equations, to get required values.

The given system of equations is :-

$\left\{\begin{array}{l}{x}^2+y^2=25\\ 2x+y=10\end{array}\right.$

We have the following second equation :-

$$\begin{gathered} 2x+y=10 \\ ⇒y=10-2x \end{gathered}$$

Put this value of $y=10-2x$in first equation, then we have :-

role="math" localid="1645357389495" $$\begin{gathered} x^2+{(10-2x)}^2=25 \\ ⇒x^2+100-40x+4x^2-25=0 \\ ⇒5x^2-40x+75=0 \\ ⇒x^2-8x+15=0 \end{gathered}$$

We have the following quadratic equation :-

$$\begin{gathered} x^2-8x+15=0 \\ ⇒x^2-3x-5x+15=0 \\ ⇒x(x-3)-5(x-3)=0 \\ ⇒(x-3)(x-5)=0 \\ ⇒x-3=0orx-5=0 \\ ⇒x=3orx=5 \\ ⇒x=3,5 \end{gathered}$$

We find :-

$x=3,5$

Also we have :-

$y=10-2x$.

So that :-

When $x=3$,

$y=10-(2×3)=10-6=4$. So solution pair is $(3,4)$.

When $x=5$,

$y=10-(2×5)=10-10=0$. So solution pair is$\left(5,0\right)$.