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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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The techniques for solving a system of linear equations in three variables are similar to those used on systems of linear equations in two variables. We eliminate variables by either substitution or addition.

Example 1

A linear system with a single solution

Solve the system:

(1) x + y - z = -1

(2) 2x - 2y + 3z = 8

(3) 2x - y + 2z = 9

Solution

We can eliminate z from Eqs. (1) and (2) by multiplying Eq. (1) by 3 and adding it to Eq. (2):

 3x + 3y - 3z = -3 Eq. (1) multiplied by 1 2x - 2y + 3z = 8 Eq.(2) (4) 5x + y = 5

Now we must eliminate the same variable, z, from another pair of equations. Eliminate z from (1) and (3):

 2x + 2y - 2z = -2 Eq. (1) multiplied by 2 2x - y + 2z = 9 Eq. (3) (5) 4x + y = 7

Equations (4) and (5) give us a system with two variables. We now solve this system. Eliminate y by multiplying Eq. (5) by -1 and adding the equations:

 5x + y = 5 Eq. (4) -4x - y = -7 Eq. (5) multiplied by -1 x = -2

Now that we have x, we can replace x by -2 in Eq. (5) to find y:

 4x + y 4(-2) + y  -8 + y y = 7= 7 = 7 = 15

Now replace x by -2 and y by 15 in Eq. (1) to find z:

 x + y - z-2 + 15 - z 13 - z -z  z = -1= -1 = -1 = -14 = 14

Check that (-2, 15, 14) satisfies all three of the original equations. The solution set is {(-2, 15, 14)}. 