Answer
$\left\{\left(x,\dfrac{1-7x}{7},\dfrac{18-7x}{7}\right)|x\text{ is any real number}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-y+2z=5\\
3x+4y-z=-2\\
5x+2y+3z=8
\end{cases}$
Multiply the second equation by 2 and add it to the first equation to eliminate $z$; then multiply the second equation by 3 and add it to the third equation to eliminate $z$:
$\begin{cases}
x-y+2z+2(3x+4y-z)=5+2(-2)\\
5x+2y+3z+3(3x+4y-z)=8+3(-2)
\end{cases}$
$\begin{cases}
x-y+2z+6x+8y-2z=5-4\\
5x+2y+3z+9x+12y-3z=8-6
\end{cases}$
$\begin{cases}
7x+7y=1\\
14x+14y=2
\end{cases}$
Multiply the first equation by -2 and add it to the second:
$-2(7x+7y)+14x+14y=-2(1)+2$
$-14x-14y+14x+14y=0$
$0=0$
We got an identity; therefore the system has infinitely many solutions.
Express $y$ in terms of $x$:
$x+y=\dfrac{1}{7}$
$y=\dfrac{1}{7}-x$
$y=\dfrac{1-7x}{7}$
Express $z$ in terms of $x$:
$3x+4y-z=-2$
$3x+4\cdot \dfrac{1-7x}{7}-z=-2$
$z=3x+2+\dfrac{4(1-7x)}{7}$
$z=\dfrac{21x+14+4-28x}{7}$
$z=\dfrac{18-7x}{7}$
The solutions set is:
$\left\{\left(x,\dfrac{1-7x}{7},\dfrac{18-7x}{7}\right)|x\text{ is any real number}\right\}$
