How do solve the following linear system?:  y = 8 x - 3 , x-3y=-1 ?

Dec 14, 2015

$x = \frac{10}{23} \mathmr{and} y = \frac{11}{23}$

Explanation:

Substitute the first equation into the second - that is, in the second equation, wherever you see a $y$ you replace it with $8 x - 3$.
Then the second equation becomes :
$x - 3 \left(8 x - 3\right) = - 1$.

This is now 1 equation in 1 unknown and we may solve for $x$ to obtain
$x - 24 x + 9 = - 1$
$\therefore x = - \frac{10}{-} 23 = \frac{10}{23}$.

Now substitute this value back into equation 1 to find $y$ :

$\therefore y = 8 \left(\frac{10}{23}\right) - 3$
$= \frac{11}{23}$