Precalculus, Chapter 7, 7.3, Section 7.3, Problem 30

Eq. 1 : 5x-3y+2z=3
Eq. 2 : 2x+4y-z=2
Eq. 3 : x+y-z=-1
Multiply Eq. 2 by 2 and add Eq. 1,
4x+8y-2z=4
5x-3y+2z=3
Eq.4 : 9x+5y=7
Subtract Eq. 3 from Eq. 2,
Eq.5 : x+3y=3
Now let's solve Eq.4 and Eq.5 by substitution method,
From Eq.5 ,
x=3-3y
Substitute the above expression of x in the Eq.4,
9(3-3y)+5y=7
27-27y+5y=7
27-22y=7
-22y=7-27
-22y=-20
y=(-20)/-22
y=10/11
Plug in the value of y in the expression of x,
x=3-3y
x=3-3(10/11)
x=3-30/11
x=(33-30)/11
x=3/11
Plug in the values of x and y in the Eq.3,
3/11+10/11-z=-1
13/11-z=-1
-z=-1-13/11
-z=(-11-13)/11
-z=(-24)/11
z=24/11
Solutions of the equations are x=3/11 , y=10/11 and z=24/11