Chapter 8.2, Problem 62E

Solution

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

The value of an equation is $(-3,0),\text{}(0,1)$ .

Given information: The given system of equations is:

  $\begin{array}{l}\frac{x^2}{9}+y^2=1\mathrm{.........}(1)\\ x-3y=-3\mathrm{..........}(2)\end{array}$

Calculation:

The equation can be converted as follows:

  $\begin{array}{l}\frac{x^2+9y^2}{9}=1\\ x^2+9y^2=9\mathrm{..........}(3)\end{array}$

Taking out the value of $x$ from equation (2),

  $x=-3+3y$

Substitute the value of $x$ in equation (3) and we get,

  $\begin{array}{c}$ {(3y-3)}^2+9y^2=9$9y^2+9-18y+9y^2=9\\ 9y^2+9-18y+9y^2-9=0\\ 18y^2-18y=0\\ 18y(y-1)=0\\ y=0,1\end{array}$

Here,

If $y=0$ , then $x=3+3(0)=-3$ .

If $y=1$ , then $x=-3+3(1)=-3+3=0$ .

The values are $(-3,0),\text{}(0,1)$ .

Thus, the system of algebraically equation is $(-3,0),\text{}(0,1)$ .

Here, the graphical representation.

  

Graphical Interpretation:

In the graph the red curve represents $\frac{x^2}{9}+y^2=1$ and green curve represents $x-3y=-3$ . And from the graph it is observed that the curves are intersecting at 2 points, $(-3,0)$ and $\text{}(0,1)$ .