Chapter 11.6, Problem 49AYU

To determine

To solve: The given system of equation.

Answer to Problem 49AYU

Solution:

$\left(0, 1\right), \left(0, -2\right), \left(2, -1\right).$

Explanation of Solution

Given:

$y^2 + y + x^2 - x - 2 = 0$   -----(Eq 1)

$y + 1 + \frac{x - 2}{y} = 0$   -----(Eq 2)

Calculation:

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

$- x + 2 + x^2 - x - 2 = 0$

The above equation can be rewritten as

$x\left(x - 2\right) = 0$

$x = 0 or 2$

Substituting $x$ in Eq 2, we get $y = 1, -2 or 0, -1$.

But $y\ne 0$ (by (Eq 2)). Therefore $y = 1, -2 or - 1.$

The solutions to the equations are $\left(0, 1\right), \left(0, -2\right), \left(2, -1\right)$.