Chapter 10.8, Problem 22E

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 22E

  $\begin{array}{c}x=4\text{ and }y=0.5\\ x=-2.8\text{ and }y=-2.9\end{array}$

Explanation of Solution

Given information:

  $x^2=16-y^2$..........(1)

  $2y-x+3=0$..........(2)

Calculation:

Solve equation (2) as follows:

  $\begin{array}{c}2y-x+3=0\\ x=2y+3\left[\text{Add }x\text{ on each side}\right]\mathrm{...}\left(3\right)\end{array}$

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

  $\begin{array}{c}{\left(2y+3\right)}^2=16-y^2\\ 4y^2+9+12y=16-y^2\\ 4y^2+y^2+12y+9-16=0\\ 5y^2+12y-7=0\\ y=\frac{-12\pm \sqrt{{12}^2-4\left(5\right)\left(-7\right)}}{2\left(5\right)}\left[\text{Using }\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ =\frac{-12\pm \sqrt{{12}^2+140}}{10}\\ =\frac{-12\pm 16.9}{10}\\ =0.5\text{ or }-2.9\end{array}$

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

  $\begin{array}{c}x=2\left(0.5\right)+3\left[\text{For }y=0.5\right]\\ =1+3\\ =4\\ x=2\left(-2.9\right)+3\left[\text{For }y=-2.9\right]\\ =-5.8+3\\ =-2.8\end{array}$

Thus, the solutions $\left(x,y\right)$ are $\left(4,0.5\right)\text{ and }\left(-2.8,-2.9\right)$ .

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. $\left(3.97,0.485\right)\text{ and }\left(-2.77,-2.885\right)$ which are same to the calculated solutions after rounding off to nearest tenth. Thus, the solutions are verified.