Answer
$(1,1,-1)$
Work Step by Step
Given Equations are
$x+y+z=1$ Equation $(1)$
$2x-y+z=0$ Equation $(2)$
$-x+2y+2z=-1$ Equation $(3)$
Adding Equation $(1)$ and Equation $(2)$
$x+y+z+2x-y+z=1+0$
$3x+2z=1$ Equation $(4)$
Adding Equation $(1)$ and Equation $(3)$
$x+y+z-x+2y+2z=1-1$
$3y+3z=0$
$y+z=0$
$y=-z$ Equation $(5)$
Substituting $y=-z$ in Equation $(1)$ we get,
$x+y+z=1$
$x-z+z=1$
$x=1$
Substituting $x$ value in Equation $(4)$ we get,
$3x+2z=1$
$3(1)+2z=1$
$3+2z=1$
$2z=1-3$
$2z=-2$
$z=-1$
Substituting $z$ value in Equation $(5)$
$y=-z$
$y=-(-1)$
$y=1$
Solution $(1,1,-1)$
Substituting $x,y,z$ values in
$\frac{x}{8} + \frac{y}{4} + \frac{z}{3} = \frac{1}{24}$
$\frac{1}{8} + \frac{1}{4} - \frac{1}{3} = \frac{1}{24}$
$\frac{3+6-8}{24} = \frac{1}{24}$
$\frac{9-8}{24} = \frac{1}{24}$
$\frac{1}{24} = \frac{1}{24}$
