# Question #58346

Sep 17, 2016

$x = \frac{9}{2} + \frac{\sqrt{17}}{2} , y = - \frac{9}{2} + \frac{\sqrt{17}}{2}$,

and,

$x = \frac{9}{2} - \frac{\sqrt{17}}{2} , y = - \frac{9}{2} - \frac{\sqrt{17}}{2}$.

#### Explanation:

Let ${\left(x - y\right)}^{\frac{1}{2}} = a \therefore \left(x - y\right) = {a}^{2}$. Using these in the ${1}^{s t}$ eqn., we

get, $3 {a}^{2} + a - 30 = 0 \Rightarrow a = - \frac{10}{3} , \mathmr{and} , a = 3$

$\therefore {\left(x - y\right)}^{\frac{1}{2}} = - \frac{10}{3} \text{ is not admissible, since, } {\left(x - y\right)}^{\frac{1}{2}} > 0.$

$\therefore {\left(x - y\right)}^{\frac{1}{2}} = 3 \Rightarrow x - y = 9 , \mathmr{and} , x = y + 9$.

By the ${2}^{n d}$ eqn., then,

$\left(y + 9\right) y + 27 = 11 , \mathmr{and} , {y}^{2} + 9 y + 16 = 0$

$\therefore y = \frac{- 9 \pm \sqrt{81 - 64}}{2} = \frac{- 9 \pm \sqrt{17}}{2} = - \frac{9}{2} \pm \frac{\sqrt{17}}{2}$.

Corresponding $x = y + 9 = - \frac{9}{2} \pm \frac{\sqrt{17}}{2} + 9 = \frac{9}{2} \pm \frac{\sqrt{17}}{2}$.

Thus, $x = \frac{9}{2} + \frac{\sqrt{17}}{2} , y = - \frac{9}{2} + \frac{\sqrt{17}}{2}$, and,

$x = \frac{9}{2} - \frac{\sqrt{17}}{2} , y = - \frac{9}{2} - \frac{\sqrt{17}}{2}$.

These roots satisfy the eqns. Hence, the soln.