# Question #e9a6e

Oct 19, 2016

$\left(x , y\right) \in \left\{\begin{matrix}2 & 4 \\ 10 & 28\end{matrix}\right\}$

#### Explanation:

From the second equation:

$3 x - y = 2$

$\implies - y = 2 - 3 x$

$\implies y = 3 x - 2$

Substituting this into the first equation:

$8 {x}^{2} - {y}^{2} = 16$

$\implies 8 {x}^{2} - {\left(3 x - 2\right)}^{2} = 16$

$\implies 8 {x}^{2} - \left(9 {x}^{2} - 12 x + 4\right) = 16$

$\implies - {x}^{2} + 12 x - 4 = 16$

$\implies {x}^{2} - 12 x + 20 = 0$

$\implies \left(x - 2\right) \left(x - 10\right) = 0$

$\implies x - 2 = 0 \mathmr{and} x - 10 = 0$

$\implies x = 2 \mathmr{and} x = 10$

Substituting these back into $y = 3 x - 2$:

$y = 3 \left(2\right) - 2 \mathmr{and} y = 3 \left(10\right) - 2$

$\implies y = 4 \mathmr{and} y = 28$

So we get $\left(x , y\right) = \left(2 , 4\right)$ or $\left(x , y\right) = \left(10 , 28\right)$.

Checking our results, we find that both $\left(x , y\right)$ pairs fulfill the given system of equations, and thus we have the solution set

$\left(x , y\right) \in \left\{\begin{matrix}2 & 4 \\ 10 & 28\end{matrix}\right\}$