Chapter 8.3, Problem 60E

Solution

To solve: The system of equations algebraically and graphically both.

The value of an equation are $\left(2\sqrt{2},1\right)\left(-2\sqrt{2},1\right)\left(-2\sqrt{2},-1\right)\left(2\sqrt{2},-1\right)$ .

Given information: The given equation is:

  $\begin{array}{l}{\frac{x}{4}}^2-y^2=1\mathrm{.....}(1)\\ x^2+y^2=9\mathrm{......}(2)\end{array}$

Calculation:

The equation (1) can be written as:

  $\begin{array}{l}{\frac{x}{4}}^2-y^2=1\\ x^2-4y^2=4\mathrm{.........}(3)\end{array}$

From equation (2) we get the value of $x^2$ ,

  $x^2=9-y^2$

Then, substitute the value of equation (2) in equation (3), we get,

  $\begin{array}{c}9-y^2-4y^2=4\\ -5y^2=4-9\\ -5y^2=-5\\ y^2=1\\ y=\pm 1\end{array}$

If the value of $y^2=1$ , then the value of $x^2$ ,

  $\begin{array}{l}{x}^2=9-1\\ x^2=8\\ x=\pm 2\sqrt{2}\end{array}$

Thus, the system of equation is $\left(2\sqrt{2},1\right)\left(-2\sqrt{2},1\right)\left(-2\sqrt{2},-1\right)\left(2\sqrt{2},-1\right)$ .

By using the graphical representation