Chapter 10.8, Problem 14E

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 14E

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

Explanation of Solution

Given information:

  $xy=2$..........(1)

  $x^2=3+y^2$..........(2)

Calculation:

Solve equation (1) as follows:

  $\begin{array}{c}xy=2\\ x=\frac{2}{y}\left[\text{Divide each side by }y\right]\mathrm{...}\left(3\right)\end{array}$

Substitute the value of $x$ in equation (2) and solve $y$ as follows:

  $\begin{array}{c}{\left(\frac{2}{y}\right)}^2=3+y^2\\ \frac{4}{y^2}=3+y^2\\ 4=y^4+3y^2\\ y^4+3y^2-4=0\end{array}$

Let $y^2=t$ and solve $y$ as shown below.

  $\begin{array}{c}{t}^2+3t-4=0\\ t=\frac{-\left(3\right)\pm \sqrt{3^2-4\left(1\right)\left(-4\right)}}{2\left(1\right)}\left[\text{Using }\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ =\frac{-3\pm \sqrt{25}}{2}\\ =\frac{-3\pm 5}{2}\\ t=-4\text{ or 1}\\ y^2=-4\text{ or 1 }\left[\text{Replace }t\text{ by }{y}^2\right]\\ y=\pm 1\left[y\ne \sqrt{-4}\right]\end{array}$

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

  $\begin{array}{c}x=\frac{2}{-1}\left[\text{For }y=-1\right]\\ =-2\\ x=\frac{2}{1}\left[\text{For }y=1\right]\\ =2\end{array}$

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

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. $\left(-2,-1\right)\text{ and }\left(2,1\right)$ which are same as the calculated solutions. Thus, the solutions are verified.