## Precalculus (6th Edition) Blitzer

The solution set of the equations is $\left\{ \left( -2,1 \right),\left( -1,2 \right) \right\}$.
Consider the pair of equations, \left\{ \begin{align} & x-y=-3 \\ & {{x}^{2}}+{{y}^{2}}=5 \end{align} \right. Consider $x-y=-3$ and solve for x in terms of y. $x=y-3$ Now, substitute the value of x in the equation ${{x}^{2}}+{{y}^{2}}=5$. ${{\left( y-3 \right)}^{2}}+{{y}^{2}}=5$ Factorize the equation using mathematical operations as follows: \begin{align} & {{y}^{2}}+9-6y+{{y}^{2}}=5 \\ & 2{{y}^{2}}-6y+4=0 \\ & {{y}^{2}}-3y+2=0 \\ & {{y}^{2}}-2y-y+2=0 \end{align} Solve this further as, $\left( y-2 \right)\left( y-1 \right)=0$ Put the factors $\left( y-2 \right),\left( y-1 \right)$ equal to zero and solve for y: \begin{align} & \left( y-2 \right)=0 \\ & y=2 \end{align} Or \begin{align} & \left( y-1 \right)=0 \\ & y=1 \end{align} Substitute the value of $y=2,y=1$ in the equation $x=y-3$ to obtain the values of x. For $y=2$ \begin{align} & x=2-3 \\ & =-1 \end{align} For $y=1$ \begin{align} & x=1-3 \\ & =-2 \end{align} From here, the solution set is $\left( -2,1 \right)$ and $\left( -1,2 \right)$ Now substitute the values of x and y obtained in the equations \left\{ \begin{align} & x-y=-3 \\ & {{x}^{2}}+{{y}^{2}}=5 \end{align} \right. to check if they satisfy the equation. Substitute $\left( -2,1 \right)$ in $x-y=-3$ \begin{align} -2-\left( 1 \right)\overset{?}{\mathop{=}}\,-3 & \\ -3=-3 & \\ \end{align} Which is true. So, $\left( -2,1 \right)$ satisfy the equation $x-y=-3$ Substitute $\left( -2,1 \right)$ in ${{x}^{2}}+{{y}^{2}}=5$ \begin{align} {{\left( -2 \right)}^{2}}+{{1}^{2}}\overset{?}{\mathop{=}}\,5 & \\ 4+1\overset{?}{\mathop{=}}\,5 & \\ 5=5 & \\ \end{align} Which is true. So, $\left( -2,1 \right)$ satisfy the equation ${{x}^{2}}+{{y}^{2}}=5$ Substitute $\left( -1,2 \right)$ in $x-y=-3$ \begin{align} -1-2\overset{?}{\mathop{=}}\,-3 & \\ -3=-3 & \\ \end{align} Which is true. So, $\left( -1,2 \right)$ satisfy the equation $x-y=-3$ Substitute $\left( -1,2 \right)$ in ${{x}^{2}}+{{y}^{2}}=5$ \begin{align} {{\left( -1 \right)}^{2}}+{{2}^{2}}\overset{?}{\mathop{=}}\,5 & \\ 1+4\overset{?}{\mathop{=}}\,5 & \\ 5=5 & \\ \end{align} Which is true. So, $\left( -1,2 \right)$ satisfy the equation ${{x}^{2}}+{{y}^{2}}=5$ Therefore, the solution set of the system of linear equations is $\left\{ \left( -2,1 \right),\left( -1,2 \right) \right\}$.