Chapter 8.3, Problem 59E

Solution

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

The value of the system of equations is $\left(4,3\sqrt{3}\right)\left(-2,0\right)$ .

Given information: The given equations are:

  $\begin{array}{c}{\frac{x}{4}}^2-{\frac{y}{9}}^2=4\mathrm{.........}(1)\\ x-\frac{2\sqrt{3}}{3}y=-2\mathrm{........}(2)\end{array}$

Calculation:

The equation (1) can be converted as we get,

  $\begin{array}{l}\frac{9x^2-4y^2}{36}=1\\ 9x^2-4y^2=36\mathrm{.........}(3)\end{array}$

Using the value of $x$ from equation (2),

Here,

  $x=-2+\frac{2\sqrt{3}}{3}y$

Put the value of $x$ in the equation.(3),

  $\begin{array}{c}9{\left(-2+\frac{2\sqrt{3}}{3}y\right)}^2-4y^2=36\\ 9\left(4+\frac{12}{9}{y}^2-\frac{8\sqrt{3}}{3}y\right)-4y^2=36\\ 36+12y^2-24\sqrt{3}y-4y^2=36\\ 8y^2-24\sqrt{3y}=0\\ 8y(y-3\sqrt{3})=0\\ y=0\\ y=3\sqrt{3}\end{array}$

If the value of $y=0,$ then

  $\begin{array}{l}x=-2+\frac{2\sqrt{3}}{3}\times 0\\ x=-2+0\\ x=-2\\ \end{array}$

The point will be $\left(-2,0\right)$ .

If the value of $y=3\sqrt{3},$ then

  $\begin{array}{l}x=-2+\frac{2\sqrt{3}}{3}\times 3\\ x=-2+2\times 3\\ x=-2+6\\ x=4\end{array}$

The point will be $\left(4,3\sqrt{3}\right)$ .

Thus, the system of equation is $\left(4,3\sqrt{3}\right)\left(-2,0\right)$ .

By using the graphical representation