## Precalculus (6th Edition) Blitzer

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