## Nov 9, 2009

### Linear Equations in Three Variables

To solve a systems of linear equations in three variables x, y, and z, we need to use the following steps:
1) Rewrite the systems of equations in three variables as systems of linear equations in two variables. Consider any two equations from the given set of three equations and eliminate one variable from those two equations.
Let us say we are eliminating the variable z. Then we will get an equation with the variables x and y and name this equation as (4). Consider another set of two equations and eliminate the same variable z. We will get another equation with the variables x and y and name this equation as (5).
2) Now, solve the two resulting equations (4) and (5) and find the value of x and y.
3) Substitute the value of x and y in any one of the three given equations and find the value of z.
4) So, the solution of the given system of equations is (x, y, z).

Let us solve an example to get clear idea of this process.
Example 1:
Solve the following systems of linear equations in three variables.
2x + 3y + 6z = 13
3x + 4y + 2z = 11
x - 3y + 5z = -5
Solution:
Step 1:
The given system contains three linear equations in three variables.
2x + 3y + 6z = 13 ----------------- (1)
3x + 4y + 2z = 11 ----------------- (2)
x - 3y + 5z = -5   ------------------ (3)
Step 2:
Consider the two equations (1) and (3).
Observe that the coefficients of y are 3 and -3 (additive inverses).
So, we can use elimination using addition method to eliminate the variable y.
2x + 3y + 6z = 13 ----------------- (1)
(+)  x - 3y + 5z = -5   ----------------- (3)
------------------------------------------------
3x         + 11z = 8
Now, we the resultant equation has only two variables, x and z. Name the equation as (4).
3x + 11z = 8 ----------------- (4)
Step 3:
Consider another set of two equations, (2) and (3).
Here, we need to eliminate the same variable y.
3x + 4y + 2z = 11 ----------------- (2)
x - 3y + 5z = -5    ----------------- (3)
Use the elimination using multiplication to eliminate the variable y.
Multiply the equation (2) by 3 and multiply the equation (3) by 4 and add them.
9x + 12y + 6z = 33
(+) 4x - 12y + 20z = -20
--------------------------------
13x          + 26z = 13
We can simplify the above equation by dividing 13 on both the sides.
x + 2z = 1
So, the resultant equation also has only two variables, x and z. Name the equation as (5).
x + 2z = 1 -------------------- (5)
Step 4:
Now, solve the two equations (4) and (5) and find the value of x and z.
3x + 11z = 8 ----------------- (4)
x + 2z = 1 -------------------- (5)
Observe that the coefficient of x is 1 in the second equation.
So, we can use substitution method to solve these two linear equations.
Solve equation (5) for x. So, subtract 2z from both the sides and simplify to isolate x.
x + 2z - 2z = 1 - 2z
x = 1 - 2z
Now, substitute the value of x in equation (4).
3(1 - 2z) + 11z = 8
Multiply 3 and (1 - 2z) and simplify.
3 - 6z + 11z = 8
3 + 5z = 8
Subtract 3 from both the sides and simplify.
3 + 5z – 3 = 8 – 3
5z = 5
Divide by 5 on both the sides and simplify to isolate z.
z = 1
Now, replace the value of z by 1 in equation (5).
x + 2(1) = 1
x + 2 = 1
Subtract 2 from both the sides to isolate x and simplify.
x + 2 - 2 = 1 - 2
x = -1
So, we have found the value of the variables x and z as -1 and 1, respectively.
Step 5:
Substitute the value of x and z in the any one of the given three linear equations to find the value of y. Let us substitute in equation (3).
x - 3y + 5z = -5
Replace x by -1 and z by 1 in the above equation and simplify.
-1 - 3y + 5(1) = -5
-1 - 3y + 5 = -5
-3y + 4 = -5
Subtract 4 from each side of the equation.
-3y + 4 - 4 = -5 - 4
-3y = -9
Divide by -3 on both the sides.
y = 3
Step 6:
So, the solution of the given system of equations is (-1, 3, 1).

Practice Questions:
Solve the following systems of linear equations in three variables.
1) 2x + y – z = -1; -3x + 2y + 2z = 9; x + y – z = 0

2) 6x + 3y + 2z = 2; -7x – 5y + 3z = 22; x + 2y – 5z = -24