To solve any system of linear equations, you will need to have at least the same amount of equations as the amount of variables. This can then be done via the algebraic method or the algebraic elimination method. You can apply either method enough times to reduce each relation from a system of three or more equations with 3 or more variables to a system of 2 equations and two variables. 

Sample: Solve for x, y and z.

2x - y + 6z = 1 (equation 1)
x - y + z = 2 (equation 2)
x + y + z = 1 (equation 3)

Steps:

1) Solve equation 1 for y.

After applying algebra, equation 1 becomes y = 2x + 6z - 1.

2) Plug the value of y or (2x + 6z -1) into equations 2 and 3. 

Equation 2:

x - y + z = 2

x -(2x + 6z -1) + z = 2

x -2x - 6z + 1 = 2

-x - 6z = 2-1

-x -6z = 1

Equation 3:

x + y + z = 1

x + 2x + 6z -1 + z = 1

3x + 7z -1 = 1

3x + 7z = -1+1

3x + 7z = 0

We now have come down to a system of two equations and two variables.

Here are the two equations and two variables:

-x - 6z = 1 (equation A)
3x + 7z = 0 (equation B)

3) Solve equation A for x.

-x - 6z = 1
-x = 6z + 1
x = -6z - 1

4) Plug the value for x into equation B to solve for z.

3x + 7z = 0

3(-6z - 1) + 7z = 0

-18z - 3 + 7z = 0

-11z - 3 = 0

-11z = 3

z = - 11/3

5) Now plug the value for z into equation A to find x.

-x - 6z = 1

-x -6(-11/3) = 1

-x + 22 = 1

-x = -22 + 1

-x = -21

x = 21

Lastly or step 6, go back to ANY of the original 3 equations to find the value of y by substituting what you found for x and z. I will select equation 3. 

Equation 3:
x + y + z = 1

21 + y -11/3 = 1

52/3 + y = 1

y = -52/3 + 1

y = -49/3

Final answer: x = 21, y = -49/3 and z = -11/3

There are cases when NO solution to a system is possible. The solution to a system takes place where the graph will cross or meet. In light of this, we can say that parallel lines (which have equal slopes) have NO solution because parallel lines never cross, touch or meet.

These two lines never meet, and thus have no intersection: