#### Answer

The solution of the required system is $\underline{\left\{ \left( 4,-3 \right),\left( -3,4 \right) \right\}}$

#### Work Step by Step

Let us consider the system of equations:
${{x}^{2}}+{{y}^{2}}=25$ (I)
$ x+y=1$ (II)
Rewrite the equation (II) as given below:
$\begin{align}
& x+y=1 \\
& y=1-x
\end{align}$
Put the value of x in equation (I), to obtain the value of y as given below:
$\begin{align}
& {{x}^{2}}+{{\left( 1-x \right)}^{2}}=25 \\
& {{x}^{2}}+1+{{x}^{2}}-2x=25 \\
& 2{{x}^{2}}-2x-24=0 \\
& {{x}^{2}}-x-12=0
\end{align}$
Factorize and solve the equation as given below:
$\begin{align}
& {{x}^{2}}-4x+3x-12=0 \\
& x\left( x-4 \right)+3\left( x-4 \right)=0 \\
& \left( x-4 \right)\left( x+3 \right)=0
\end{align}$
Either $ x-4=0$ or $ x+3=0$
So, $ x=-3,4$
Put in the value of x in equation (II) to obtain the value of y as follows:
Put $ x=-3$:
$\begin{align}
& -3+y=1 \\
& y=4
\end{align}$
Again, put $ x=4$:
$\begin{align}
& 4+y=1 \\
& y=-3
\end{align}$
Thus, the values of the system of equations are $\left( x,y \right)=\left\{ \left( 4,-3 \right),\left( -3,4 \right) \right\}$.