Chapter 3.5, Problem 24E

To determine

To evaluate: the solution of the system by substitution.

Answer to Problem 24E

The solution of the given equations is $\left(-1,-2\right)\text{ and }\left(2,1\right)$ .

Explanation of Solution

Given:

  $\begin{array}{l}{x}^2+y^2=5\\ -x+y=-1\end{array}$

It is being asked to use method of substitution,

Finding the value of ‘ x’ from 2^nd equation,

  $x=y+1$

Substituting the value of ‘ x’ in second equation,

  $\begin{array}{l}{x}^2+y^2=5\\ {\left(y+1\right)}^2+y^2=5\\ y^2+2y+1+y^2=5\\ 2y^2+2y-4=0\end{array}$

Now, solving above quadratic equation using quadratic formula,

  $\begin{array}{c}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(2\right)\left(-4\right)}}{4}\\ =\frac{-2\pm \sqrt{4+32}}{4}\\ =\frac{-2\pm \sqrt{36}}{4}\\ =\frac{-2\pm 6}{4}\\ =-2,1\end{array}$

When $y=-2$ ,

  $\begin{array}{c}x=y+1\\ =-2+1\\ =-1\end{array}$

When

  $y=1$ ,

  $\begin{array}{c}x=y+1\\ =1+1\\ =2\end{array}$

So, the solution of the given equations is $\left(-1,-2\right)\text{ and }\left(2,1\right)$ .