Chapter 9.7, Problem 8CYU

To determine

To find: The solution for the given system of the equation.

Answer to Problem 8CYU

The solution for the equation is $\left(\sqrt{3},\sqrt{5}\right),\left(-\sqrt{3},\sqrt{5}\right),\left(\sqrt{3},-\sqrt{5}\right),\left(-\sqrt{3},-\sqrt{5}\right)$ .

Explanation of Solution

Given:

The given system of equation is,

  $\begin{array}{l}4y^2-3x^2=11\\ 3y^2+2x^2=21\end{array}$

Calculation:

Consider the given equation are,

  $\begin{array}{l}4y^2-3x^2=11\\ 3y^2+2x^2=21\end{array}$

The second equation is evaluated as,

  $\begin{array}{l}3y^2+2x^2=21\\ x^2=\frac{21}{2}-\frac{3}{2}{y}^2\end{array}$

Substitute $x^2=\frac{21}{2}-\frac{3}{2}{y}^2$ in the equation $4y^2-3x^2=11$ and solve for y as,

  $\begin{array}{l}4y^2-3\left(\frac{21}{2}-\frac{3}{2}{y}^2\right)=11\\ 8y^2-63+9y^2=22\\ y^2=5\\ y=\sqrt{5}\end{array}$

Substitute $y^2=5$ in the equation $x^2=\frac{21}{2}-\frac{3}{2}{y}^2$ and solve for $x$ as,

  $\begin{array}{l}{x}^2=\frac{21}{2}-\frac{3}{2}\left(5\right)\\ x=\pm \sqrt{3}\end{array}$

Thus, the solution for the equation is $\left(\sqrt{3},\sqrt{5}\right),\left(-\sqrt{3},\sqrt{5}\right),\left(\sqrt{3},-\sqrt{5}\right),\left(-\sqrt{3},-\sqrt{5}\right)$ .