# How do you solve xy=12 and 5x-3y=-8?

Jun 8, 2016

$y = \frac{12}{x}$

$5 x - 3 \left(\frac{12}{x}\right) = - 8$

$5 x - \frac{36}{x} = - 8$

$\frac{5 {x}^{2} - 36}{x} = - 8$

$5 {x}^{2} - 36 = - 8 x$

$5 {x}^{2} + 8 x - 36 = 0$

$5 {x}^{2} - 10 x + 18 x - 36 = 0$

$5 x \left(x - 2\right) + 18 \left(x - 2\right) = 0$

$\left(5 x + 18\right) \left(x - 2\right) = 0$

$x = - \frac{18}{5} \mathmr{and} 2$

$\therefore - \frac{18}{5} \times y = 12 \mathmr{and} 2 \times y = 12$

$y = \frac{12}{- \frac{18}{5}} \mathmr{and} y = 6$

$y = - \frac{10}{3} \mathmr{and} y = 6$

Our solution set is therefore $\left\{- \frac{18}{5} , - \frac{10}{3}\right\} \mathmr{and} \left\{2 , 6\right\}$.

Hopefully this helps!