Answer
$x_1=-\frac{3t}{2},x_2=\frac{5t}{2},x_3=t$
Work Step by Step
We have the following system:
$2x_1+4x_2-7x_3=0,..(1)\\
x_1-3x_2+9x_3=0,(2)$
Subtracting twice equation $(2)$ from $(1)$:
$(2x_1+4x_2-7x_3=0 )-2(x_1-3x_2+9x_3=0)\\
\Rightarrow 4x_2+6x_2-7x_3-18x_3=0\\
\Rightarrow 10x_2-25x_3=0\\
\Rightarrow x_2=\frac{5}{2} x_3 $
Put this in equation$(2)$:
$x_1-3\frac{5x_3}{2}+9x_3=0\\
\Rightarrow x_1+\frac{3x_3}{2}=0\\
\Rightarrow x_1=-\frac{3x_3}{2}$
This means that $x_3$ is a free variable. Therefore, let $x_3=t$. Then the solution is given by:
$x_1=-\frac{3t}{2},x_2=\frac{5t}{2},x_3=t$
