4x1 + 3x2 = 4
(2/3)x1 + 4x2 = 3
I can't seem to use back substitution
How do I solve this system?
$4x_1 + 3x_2 = 4$
$\dfrac{2}{3} x_1 + 4x_2 = 3$
from the first equation, $x_1 = 1 - \dfrac{3}{4}x_2$
substitute for $x_1$ in the second equation
$\dfrac{2}{3}\left(1 - \dfrac{3}{4}x_2\right) + 4x_2 = 3$
$\dfrac{2}{3} - \dfrac{1}{2} x_2 + 4x_2 = 3$
$\dfrac{7}{2} x_2 = \dfrac{7}{3}$
$x_2 = \dfrac{2}{3}$
back substitute to find $x_1$