1. Solve the system

4x1 + 3x2 = 4
(2/3)x1 + 4x2 = 3

I can't seem to use back substitution
How do I solve this system?

2. Re: Solve the system

Originally Posted by rhizome
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 $x_1$

3. Re: Solve the system

I don't understand your result after
(2/3)(1 - (3/4)x2) + 4x2 = 3

4. Re: Solve the system

Ah I understand now.