How do you solve the following system?:  1/9x-3/13y=1 , 3/2x - 8y = 2

Jun 8, 2017

color(blue)(y=-0.25,x=0

Explanation:

$\therefore \frac{1}{9} x - \frac{3}{13} y = 1$----(1)

$\therefore \frac{3}{2} x - 8 y = 2$----(2)

multiply $\left(2\right) \times 2$

$\therefore 3 x - 16 y = 4$

$\therefore - 16 y = - 3 x + 4$

$\therefore 16 y = 3 x - 4$

:.color(blue)(y=(3x-4)/16

substitute color(blue)(y=(3x-4)/16--- in $\left(2\right)$

$\therefore \frac{3}{2} x - 8 \left(\frac{3 x - 4}{16}\right) = 2$

$\therefore \frac{3}{2} x - \frac{24 x + 32}{16} = 2$

multiply both sides by 16

$\therefore 24 x - 24 x + 32 = 32$

$\therefore 24 x - 24 x = 32 - 32$

$\therefore 0 = 0$

:.color(blue)(x=0

substitute color(blue)(x=0 in $\left(2\right)$

$\therefore \frac{3}{2} \left(\textcolor{b l u e}{0}\right) - 8 y = 2$

$\therefore - 8 y = 2$

multiply both sides by$- 1$

$\therefore 8 y = - 2$

$\therefore y = - \frac{2}{8}$

:.color(blue)(y=-0.25 or color(blue)( -1/4

check:

substitute color(blue)(y=-0.25 in $\left(2\right)$

$\therefore \frac{3}{2} x - 8 \left(\textcolor{b l u e}{- 0.25}\right) = 2$

$\therefore \frac{3}{2} x + 2 = 2$

$\therefore \frac{3}{2} x = 2 - 2$

$\therefore \frac{3}{2} x = 0$

:.color(blue)(x=0