Answer
$\left\{\left(x,\dfrac{9x+33}{14},\dfrac{4x-39}{7},\right)|x\text{ is any real number}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x-4y+z=-15\\
x+2y-4z=27\\
5x-6y-2z=-3
\end{cases}$
Use the elimination method. Multiply the first equation by 4 and add it to the second to eliminate $z$. Then multiply the first equation by 2 and add it to the third to eliminate $z$:
$\begin{cases}
x+2y-4z+4(2x-4y+z)=27+4(-15)\\
5x-6y-2z+2(2x-4y+z)=-3+2(-15)
\end{cases}$
$\begin{cases}
x+2y-4z+8x-16y+4z=27-60\\
5x-6y-2z+4x-8y+2z=-3-30
\end{cases}$
$\begin{cases}
9x-14y=-33\\
9x-14y=-33
\end{cases}$
Multiply the first equation by -1 and add it to the second to eliminate $z$ and find $x$:
$-(9x-14y)+9x-14y=-(-33)-33$
$-9x+14y+9x-14y=33-33$
$0=0$
We got an identity; therefore the system has infinitely many solutions.
Determine $y$ in terms of $x$:
$9x-14y=-33$
$14y=9x+33$
$y=\dfrac{9x+33}{14}$
Determine $z$ in terms of $x$:
$x+2y-4z=27$
$x+2\cdot \dfrac{9x+33}{14}-4z=27$
$x+\dfrac{9x+33}{7}-4z=27$
$\dfrac{7x+9x+33}{7}-27=4z$
$z=\dfrac{16x+33-189}{28}$
$z=\dfrac{16x-156}{28}$
$z=\dfrac{4x-39}{7}$
The solution set is:
$\left\{\left(x,\dfrac{9x+33}{14},\dfrac{4x-39}{7},\right)|x\text{ is any real number}\right\}$
