Chapter 11.6, Problem 48AYU

To determine

To solve: The given system of equation.

Answer to Problem 48AYU

Solution:

$\left(3, -3\right), \left(-2, 2\right).$

Explanation of Solution

Given:

$x^2 - xy - 2y^2 = 0$   -----(Eq 1)

$xy + x + 6 = 0$   -----(Eq 2)

Calculation:

Substituting $y = \frac{- 6 - x}{x}$ in Eq 1 we get

$x^2 + x + 6 - 2{\left(\left(-6 - x\right)/x\right)}^2 = 0$

The above equation can be rewritten as

$x^4 + x^3 + 4x^2 - 24x - 72 = 0$

Further solving this equation,

$\left(x - 3\right)\left(x + 2\right)\left(x^2 + 2x + 12\right) = 0$

$x = 3 or - 2 or - 1\pm \sqrt{11}i$

The real roots are

$x = 3 or - 2.$

Substituting $x$ in Eq 2, we get $y = -3 or 2.$

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