Chapter 10.8, Problem 19E

To determine

To find: the solutions algebraically for the system of equations and then verify the answer by graphing each system. Round to the nearest tenth.

Answer to Problem 19E

  $\left(0,-1\right)\text{ and }\left(-3,2\right)$

Explanation of Solution

Given information:

  ${\left(y-1\right)}^2=4+x$..........(1)

  $x+y=-1$..........(2)

Calculation:

Solve equation (2) as follows:

  $\begin{array}{c}x+y=-1\\ x=-1-y\left[\text{Subtract }y\text{ from each side}\right]\mathrm{...}\left(3\right)\end{array}$

Substitute the value of $x$ in equation (1) and solve $y$ as follows:

  $\begin{array}{c}{\left(y-1\right)}^2=4+\left(-1-y\right)\\ y^2+1-2y=4-1-y\\ y^2-y-2=0\\ y=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-2\right)}}{2\left(1\right)}\left[\text{Using }\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ =\frac{1\pm \sqrt{9}}{2}\\ =\frac{1\pm 3}{2}\\ =2\text{ or }-1\end{array}$

Substitute each value of $y$ in equation (3) and solve $x$ as shown below.

  $\begin{array}{c}x=-1-2\left[\text{For }y=2\right]\\ =-3\\ x=-1-\left(-1\right)\left[\text{For }y=-1\right]\\ =-1+1\\ =0\end{array}$

Thus, the solutions $\left(x,y\right)$ are $\left(0,-1\right)\text{ and }\left(-3,2\right)$ .

Now use graphing calculator to plot equation (1) and (2) on same screen as shown below.

  

The solution $\left(x,y\right)$ obtained from the graph i.e. $\left(0,-1\right)\text{ and }\left(-3,2\right)$ which are same as the calculated solutions. Thus, the solutions are verified.