#### Answer

$x_{1} = -t$
$x_{2} = 2t$
$x_{3} = t$
$t$ is any real number.

#### Work Step by Step

Start by writing the equations in reduced row echelon form. We can accomplish this by interchanging Equation 1 and 2 and then row reducing.
$2x_{1} + x_{2} = 0$
$5x_{1} + 2x_{2} + x_{3} = 0$
Dividing Equation 1 by 2 gives...
$x_{1} + \frac{1}{2}x_{2} = 0$
$5x_{1} + 2x_{2} + x_{3} = 0$
Adding -5 times Equation 1 to Equation 2 will eliminate the $x_1$ variable from Equation 2.
$x_{1} + \frac{1}{2}x_{2} = 0$
$0x_{1} + -\frac{1}{2}x_{2} + x_{3} = 0$
Multiply Equation 2 by -2.
$x_{1} + \frac{1}{2}x_{2} = 0$
$0x_{1} + x_{2} -2 x_{3} = 0$
Because there 2 pivots, $x_{3}$ is a free variable. Solve both equations for $x_{1}$ and $x_{2}$ respectively.
$x_{1} = -\frac{1}{2}x_{2}$
$x_{2} = 2x_{3}$
Since $x_3$ is free, it can be any real number $t$.
$x_3 = t$
Through back substitution, we can express $x_{1}$ and $x_{2}$ in terms of $t$.
$x_{2} = 2t$
$x_{1} = -\frac{1}{2}*2t = -t$