# Question #a9fdf

Nov 6, 2016

$\left(2 , 3\right) \text{ or } \left(3 , 2\right)$

#### Explanation:

Rearrange x + 5y in terms of x or y and substitute into xy = 6.

$x + y = 5 \Rightarrow y = 5 - x$

substitute this into xy = 6

$\Rightarrow x \left(5 - x\right) = 6$

distribute and equate to zero.

$\Rightarrow 5 x - {x}^{2} = 6 \Rightarrow {x}^{2} - 5 x + 6 = 0$

$\left(x - 2\right) \left(x - 3\right) = 0 \Rightarrow x = 2 , x = 3$

$x = 2 \to y = 5 - 2 = 3 \to \left(2 , 3\right) \text{ is a solution}$

$x = 3 \to y = 5 - 3 = 2 \to \left(3 , 2\right) \text{ is also a solution}$

This can be checked fairly ' easily' as follows.

$x = 2 , y = 3 \Rightarrow x + y = 2 + 3 = 5 \text{ and } x y = 2 \times 3 = 6$

$x = 3 , y = 2 \Rightarrow x + y = 3 + 2 = 5 \text{ and } x y = 3 \times 2 = 6$