Chapter 11.6, Problem 36AYU

To determine

To solve: The given system of equation.

Answer to Problem 36AYU

Solution:

$(2\sqrt{2}, \sqrt{3}), \left(2\sqrt{2}, -\sqrt{3}\right), \left(- 2\sqrt{2}, \sqrt{3}\right), \left(- 2\sqrt{2}, -\sqrt{3}\right).$

Explanation of Solution

Given:

$x^2 - 3y^2 + 1 = 0$   -----(Eq 1)

$2x^2 - 7y^2 + 5 = 0$   -----(Eq 2)

Calculation:

Substituting $y^2 = \frac{x^2 + 1}{3}$ in Eq 1 we get

$2x^2 - 7\left(\frac{x^2 + 1}{3}\right) + 5 = 0$

The above equation can be rewritten as

$-x^2 + 8 = 0$

$x^2 = 8$

$x = \pm 2\sqrt{2}$

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

The solutions to the equations are $(2\sqrt{2}, \sqrt{3}), \left(2\sqrt{2}, -\sqrt{3}\right), \left(- 2\sqrt{2}, \sqrt{3}\right), \left(- 2\sqrt{2}, -\sqrt{3}\right).$