Precalculus, Chapter 7, 7.3, Section 7.3, Problem 32

(1) 2x+y+3z=1
(2) 2x+6y+8z=3
(3) 6x+8y+18z=5

Using Equations (1) and (2) eliminate the x. Multiply equation (2) by -1.
2x+1y+3z=1
-2x-6y-8z=-3
----------------------------
-5y-5z=-2
(4) -5y-5z=-2

Using Equations (2) and (3) eliminate the x. Multiply equation (2) by -3.
-6x-18y-24z=-9
6x+8y+18z=5
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-10y-6z=-4
(5) -10y-6z=-4

Using equations (4) and (5) eliminate the y. Multiply equation 4 by -2.
10y+10z=4
-10y-6z=-4
-----------------------
4z=0
z=0

Use equation (4) to solve for y.
-5y-5z=-2
-5y-5(0)=-2
y=2/5

Use equation (1) to solve for x.
2x+y+3z=1
2x+2/5+3(0)=1
x=3/10

The solution set is (3/10, 2/5, 0).

You may check your answers by plugging in the x, y, z values into the original equations.
(1) 2(x)+y+3z=1
2(3/10)+(2/5)+3(0)=1

6/10+2/5=1
6/10+4/10=1
1=1

(2) 2x+6y+8z=3
2(3/10)+6(2/5)+8(0)=3
6/10+12/5=3
6/10+24/10=3
30/10=3
3=3

(3) 6x+8y+18z=5
6(3/10)+8(2/5)+18(0)=5
18/10+16/5=5
18/10+32/10=5
50/10=5
5=5