Chapter 11.6, Problem 39AYU

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

$\{\begin{array}{l}2x^2\text{ + }{y}^2\text{ = }2\\ x^2-2y^2\text{ + }8\text{ = }0\end{array}$

To determine

To solve: The given system of equation.

Answer to Problem 39AYU

Solution:

No real solution exists.

Explanation of Solution

Given:

$2x^2 + y^2 = 2$   -----(Eq 1)

$x^2 - 2y^2 + 8 = 0$   -----(Eq 2)

Calculation:

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

$x^2 - 2\left(2 - 2x^2\right) + 8 = 0$

The above equation can be rewritten as

$5x^2 + 4 = 0$

$x^2 = -4/5$

$x = \pm \sqrt{}\left(\frac{4}{5}\right)i$

Therefore, no real solution exist.