Systems with Three Variables
In order to find the value of two variables, we need two equations. To find the value of three variables, we need three equations. If we have more than three variables, we need as many equations as we have variables. To find the value of 100 variables, we would need 100 equations!
We solve a system of equations with three variables just as we solve a system of equations with two variables; we just need to use more steps.

 x + y + z = 3
 3z + 2y = 4x
 5x – z = –1 – y
We start by writing one variable in terms of two other variables. We can use the first equation to write any of the variables in terms of the others. Let's write x in terms of y and z:

 x + y + z = 3
 x = 3 – y – z
Next, replace x in each of the other two equations with the expression 3 – y – z:
3z + 2y = 4(3 – y – z)  5(3 – y – z) – z = –1 – y 
3z + 2y = 12 – 4y – 4z  15 – 5y – 5z – z = –1 – y 
7z + 2y = 12 – 4y  15 – 4y – 6z = –1 
7z + 6y = 12  –4y – 6z = –16 
We now have two equations and two variables:

 7z + 6y = 12
 –4y – 6z = –16
Write one variable in terms of the other. Let's write y in terms of z using the second equation:

 –4y – 6z = –16
 –4y = 6z – 16
Now, replace y in the first equation with the expression that is equal to y,

 7z – 9z + 24 = 12
 –2z + 24 = 12
 –2z = –12
 z = 6
We have the value of one variable, z. Replace z with its value in either equation and solve for the value of y:

 7z + 6y = 12
 7(6) + 6y = 12
 42 + 6y = 12
 6y = –30
 y = –5
We have the values of y and z. Substitute them into any one of the original three equations to find the value of x:

 x + y + z = 3
 x + –5 + 6 = 3
 x + 1 = 3
 x = 2
The solution to this system of equations is x = 2, y = –5, z = 6.
Find practice problems and solutions for these concepts at Systems of Equations—Solving with Substitution Practice Questions.
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