Precalculus, Chapter 7, 7.3, Section 7.3, Problem 11

2x-y+5z=24
y+2z=6
z=8
Substitute back the value of z in the second equation,
y+2*8=6
y+16=6
y=6-16
y=-10
Now substitute back the value of y and z in the first equation,
2x-(-10)+5*8=24
2x+10+40=24
2x+50=24
2x=24-50
2x=-26
x=(-26)/2
x=-13
Solutions of the equations are x=-13 , y=-10 and z=8