Chapter 11.6, Problem 41AYU

In Problems 25-54, solve each system. Use any method you wish.

$\{\begin{array}{l}{x}^2\text{ + 2}{y}^2\text{ = }16\\ 4x^2-y^2\text{ = }24\end{array}$

To determine

To solve: The given system of equation.

Answer to Problem 41AYU

Solution:

$\left(\frac{8}{3}, \frac{2\sqrt{10}}{3}\right), \left(\frac{8}{3}, -\frac{2\sqrt{10}}{3}\right), \left(-\frac{8}{3}, \frac{2\sqrt{10}}{3}\right), \left(-\frac{8}{3}, -\frac{2\sqrt{10}}{3}\right)$

Explanation of Solution

Given:

$x^2 + 2y^2 = 16$   -----(Eq 1)

$4x^2 - y^2 = 24$   -----(Eq 2)

Calculation:

Substituting $y^2 = 4x^2 - 24$ in Eq 1 we get

$x^2 + 2\left(4x^2 - 24\right) = 16$

The above equation can be rewritten as

$9x^2 - 64 = 0$

$x^2 = 64/9$

$x = \pm 8/3$

Substituting $x$ in Eq 2, we get $y = \pm \frac{2\sqrt{10}}{3}, or\pm \frac{2\sqrt{10}}{3}$.

The solutions to the equations are

$\left(\frac{8}{3}, \frac{2\sqrt{10}}{3}\right), \left(\frac{8}{3}, -\frac{2\sqrt{10}}{3}\right), \left(-\frac{8}{3}, \frac{2\sqrt{10}}{3}\right), \left(-\frac{8}{3}, -\frac{2\sqrt{10}}{3}\right)$