Answer
$(-6, 12, -4)$
Work Step by Step
$\frac{1}{3}x-\frac{1}{4}y+z=-9$ Equation $(1)$
$\frac{1}{2}x-\frac{1}{3}y-\frac{1}{4}z=-6$ Equation $(2)$
$x-\frac{1}{2}y-z=-8$ Equation $(3)$
Equation $(1)$ + Equation $(3)$
$(\frac{1}{3}x-\frac{1}{4}y+z) + (x-\frac{1}{2}y-z) = -9+(-8)$
$\frac{4}{3}x-\frac{3}{4}y=-17$ Equation $(4)$
$4\times$Equation $(2) - $Equation $(3)$
$4(\frac{1}{2}x-\frac{1}{3}y-\frac{1}{4}z)-(x-\frac{1}{2}y-z)=4(-6)-(-8)$
$(2x-\frac{4}{3}y-z)-(x-\frac{1}{2}y-z)=-24+8$
$x-\frac{5}{6}y=-16$ Equation $(5)$
$\frac{4}{3}\times$ Equation $(5) - $Equation $(4)$
$\frac{4}{3}(x-\frac{5}{6}y)-(\frac{4}{3}x-\frac{3}{4}y)=\frac{4}{3}(-16)-(-17)$
$(\frac{4}{3}x-\frac{10}{9}y)-(\frac{4}{3}x-\frac{3}{4}y)=-\frac{64}{3}-(-17)$
$\frac{13}{36}y=\frac{13}{3}$
$y=12$
Substitute $y=12$ into Equation $(5)$ to get $x$.
$x-(\frac{5}{6}\times 12)=-16$
$x-10=-16$
$x=-6$
Substitute known values for $x$ and $y$ into Equation $(3)$ to get $z$.
$-6-(\frac{1}{2}\times12)-z=-8$
$-6-6-z=-8$
$-12-z=-8$
$-z=4$
$z=-4$
$(-6, 12, -4)$ satisfies the given equations.
Solution is $(-6, 12, -4)$.
