Linear Equations with Three Variables

We will now learn how to solve linear equations in three variables. The system of linear equations is generally in the form a1x+b1 y+c1z = d1 a2x+b2 y+c2z = d2 a3x+b3 y+c3z = d3 To solve these equations, we select a variable, either x or y or z (usually z is chosen), and eliminate it from the system of equations. We then obtain simultaneous linear equations in two variables x and y, which we solve as we did in section 2.3. After obtaining the solutions to x and y, we substitute these values in any one of the original equations to obtain the value of z. Consider these examples.

Example 1

Solve the linear equations. x+2y+2z=11         ----------- (1) 2x+y+z=7             ------------ (2) 3x+4y+z=14         ------------ (3) To eliminate z from (1), (2) and (3), we multiply equations 2 and 3 by 2. 2∗ (2x+y+z=7) 2∗ (3x+4y+z=14)    4x+2y+2z=14        ------- (4)    6z+8y+2z=28        ------- (5) Subtracting 1 from 4 and 1 from 5, we get 4x+ 2y+ 2z=14 -x ± 2y ± 2z=-11 ____________          3x= 3     ∴    x=3/3            x=1 6x+ 8y + 2z =28 -x ± 2y ± 2z =-11 ___________    5x+6y = 17 Substitute x=1 in this equation.            5∗1+6y = 17            6y =17-5            6y =12              y =12/6              y =2       Substitute x=1, y=2 in equation 1.    x+2y+2z = 11 1+2 ∗2+2z = 11      1+4+2z = 11              2z = 11-5              2z = 6                z = 6/2                z = 3 Solution set = {(1,2,3)}.

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