Chapter 11.6, Problem 32AYU

To determine

To solve: The given system of equation.

Answer to Problem 32AYU

Solution:

$\left(- 2, 0\right), \left(7, 6\right)$

Explanation of Solution

Given:

$2y^2 - 3xy + 6y + 2x + 4 = 0$   -----(Eq 1)

$2x - 3y + 4 = 0$   -----(Eq 2)

Calculation:

Substituting $x = \frac{3y - 4}{2}$ in Eq 1 we get

$2y^2 - 3y\left(\frac{3y - 4}{2}\right) + 6y + 3y = 0$

The above equation can be rewritten as

$-\left(\frac{5}{2}\right)y^2 + 15y = 0$

$y = 0 or 6$

Substituting $y$ in Eq 2, we get $x = - 2$ or 7

The solutions to the equations are $\left(- 2, 0\right), \left(7,6\right).$