Chapter 8.2, Problem 61E

Solution

To solve: The system of equations algebraically and graphically.

The value of a system of an equation is $(2,0),\text{}(-2,0)$ .

Given information: The given system of an equations is:

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

Calculation:

The equation (1) can be converted as:

  $\begin{array}{l}\frac{9x^2+4y^2}{36}=1\\ 9x^2+4y^2=36\end{array}$

Taking out the value of $y^2$ from equation (2), we get

  $y^2=4-x^2\mathrm{.............}(3)$

Substitute the value of equation (3) in equation (1) convertible form,

  $\begin{array}{c}9x^2+4(4-x^2)=36\\ 9x^2+16-4x^2=36\\ 5x^2+16=36\\ 5x^2=36-16\\ 5x^2=20\\ x^2=4\\ x=\pm 2\end{array}$

Put the value of $x^2$ in the equation (3):

  $\begin{array}{l}{y}^2=4-4\\ y^2=0\\ y=0\end{array}$

Thus, the value of an equations is $(2,0),\text{}(-2,0)$ .

Here, the graphical representation is as followed:

  

Graphical interpretation:

In the graph the red curve represents $\frac{x^2}{4}+\frac{y^2}{9}=1$ and green curve represents $x^2+y^2=4$

  $\frac{x^2}{4}+\frac{y^2}{9}=1$ from the graph it is observed that the curves are intersecting at 2 points, $\left(2,0\right)$ and $\left(-2,0\right)$ .