# How do you solve this equation?

## $x - 2 y = 3$ ${x}^{2} + 2 {y}^{2} = 27$ This is a simultaneous equation

Apr 28, 2018

$\left(i\right) x = - 3 + 6 \sqrt{2} \mathmr{and} y = - 3 + 3 \sqrt{2}$

$\left(i i\right) x = - 3 - 6 \sqrt{2} \mathmr{and} y = - 3 - 3 \sqrt{2}$

#### Explanation:

Here,

color(red)(x-2y=3...to(I) and

color(red)(x^2+2y^2=27...to(II)

Now, x-2y=3=>color(red)(x=2y+3...to(III)

From , $\left(I I\right)$

${\left(2 y + 3\right)}^{2} + 2 {y}^{2} = 27$

$\implies 4 {y}^{2} + 12 y + 9 + 2 {y}^{2} = 27$

$\implies 2 {y}^{2} + 12 y + 9 - 27 = 0$

$\implies 2 {y}^{2} + 12 y - 18 = 0$

$\implies {y}^{2} + 6 y - 9 = 0$

Comparing with $a {x}^{2} + b x + c = 0$

$a = 1 , b = 6 , c = - 9$

$\therefore \triangle = {b}^{2} - 4 a c = 36 - 4 \left(1\right) \left(- 9\right) = 36 + 36$

$\sqrt{\triangle} = \sqrt{36 \times 2} = 6 \sqrt{2}$

$\therefore y = \frac{- b \pm \sqrt{\triangle}}{2 a} = \frac{- 6 \pm 6 \sqrt{2}}{2} = - 3 \pm 3 \sqrt{2}$

From $\left(I I I\right) ,$

$\left(i\right) y = - 3 + 3 \sqrt{2} \implies x = 2 \left(- 3 + 3 \sqrt{2}\right) + 3 = - 3 + 6 \sqrt{2}$

=>color(blue)( x=-3+6sqrt2 and y=-3+3sqrt2

$\left(i i\right) y = - 3 - 3 \sqrt{2} \implies x = 2 \left(- 3 - 3 \sqrt{2}\right) + 3 = - 3 - 6 \sqrt{2}$

=>color(blue)( x=-3-6sqrt2 and y=-3-3sqrt2