# Question #df17c

Nov 26, 2014

By raising $6$ to both sides of the equations,

$\left\{\begin{matrix}3 {x}^{2} + {y}^{2} = 28 \\ x + y = 6 \implies y = 6 - x\end{matrix}\right.$

by substituting the second equation into the first equation,

$3 {x}^{2} + {\left(6 - x\right)}^{2} = 28 \implies 4 {x}^{2} - 12 x + 8 = 0$

by dividing by $4$,

$\implies {x}^{2} - 3 x + 2 = 0$

by factoring out,

$\implies \left(x - 1\right) \left(x - 2\right) = 0 \implies x = 1 , 2 \implies y = 5 , 4$

Hence, the solutions are

$\left(x , y\right) = \left(1 , 5\right)$ and $\left(2 , 4\right)$

I hope that this was helpful.