Answer
Infinitely many solutions.
Solution set: $\{(x,y,z)|x+2y-3z=4\}$
Work Step by Step
$-2x-4y+6z=-8$ Equation $(1)$
$x+2y-3z=4$ Equation $(2)$
$4x+8y-12z=16$ Equation $(3)$
$-\frac{1}{2} \times$ Equation $(1)$
$-\frac{1}{2}(-2x-4y+6z)=-\frac{1}{2}\times-8$
$x+2y-3z=4$ This is identical to Equation $(2)$.
$\frac{1}{4} \times$ Equation $(3)$
$\frac{1}{4}(4x+8y-12z)=\frac{1}{4}\times16$
$x+2y-3z=4$ This is identical to Equation $(2)$.
Since all three equations are identical and equivalent, they are dependent equations. Therefore, the system has infinitely many solutions.
Solution set: $\{(x,y,z)|x+2y-3z=4\}$
