Solution is x=5, y=2, z=1 OR (5,2,1)

We’ll use the Gaussian elimination here (which is all about adding and subtracting equations):
Add the first equation to the second (keep the first)

Add equations (1) and (2) to eliminate z:

3x-y=13

Add 4 times equations (2) and (3) to eliminate z again:

11x-11y=33 OR x-y=3

Now solve system by substitution:

x=5 and y=2

Substitute x and y into an original equation to find z=1.

Add equations (1) and (2) to eliminate x and z:

4y=-12

y=-3

Substitute y into equations (1) and (3) to solve simpler system and find x=2 and z=1

Keep equation 3, eliminate z in eq. 2 and eq. 1
20x + 14 y = 46
-5x – 7 y = -8
3x + 3y –z = 4

equation 1 + 2(equation 2) = new equation 2

20x + 14 y = 46
10x = 30
3x + 3y –z = 4

Now find the values by filling in your answers:
x = 3, y = -1 and z = 2

Use equation (2) to eliminate x

5y – 2z = -17
- x + 3y +z = -10
11y + z = -32

put z from eq 3 in eq 1:
5y – 2( -32 – 11y) = -17 =>

y = -3
eq 1 gives: z = 1
eq 2 gives: x = 2

Keep equation 1, subtract eq.1 from both eq. 2 and eq. 3

x + y – z = 5
-y + 2z = -4
2y – 4z = 2

Now do this: 2 (eq. 2) + eq. 3 = new equation 3
x + y – z = 5
-y + 2z = -4
0y +0z = …This means there is no solution possible here.

Use eq 1 to eliminate z in eq 2 and 3:
2x + y – z = 1
x – y = 3
6x – 6y = 18 (is the same as eq2!!)
We have two equations with three unknown variables.

Take y out of equation 1 and x out of equation 2:
y = - 1.5 z + 1.5 x
x = 10 y – 6 z
x² + y² - z² = 16

put x from eq 2 in eq 1 and y from eq 1 in eq 2
y = 0.75 z
x = 1.5 z
x² + y² - z² = 16

Now put x and y in eq 3:
2.25 z² + 0.56 z² - z²= 16 => z = 3.0
So y = 2.2 and x = 4.5