Answer
$x_1=-\frac{37}{2}t,x_2=\frac{9}{2}t,x_3=t$
Work Step by Step
We are given the following system
$x_1+3x_2+5x_3=0,(1)\\
x_1+4x_2+\frac{1}{2}x_3=0,(2)$
Subtracting equation $(1)$ from $(2)$, gives
$-x_2+\frac{9}{2}x_3=0\\
\Rightarrow x_2=\frac{9}{2}x_3$
Substituting equation$(1)$, we have
$x_1+3\frac{9}{2}x_3+5x_3=0\\
\Rightarrow x_1+\frac{37}{2}x_3=0\\
\Rightarrow x_1=-\frac{37}{2}x_3$.
We have $x_3$ as the free variable; therefore, take $x_3=t$
$\Rightarrow x_1=-\frac{37}{2}t$ and $x_2=\frac{9}{2}t$.
