Sava D. answered • 06/11/20

Math master degree tutor with good understanding of precalculus

We will use addition of equations for forward step and back substitution for the solving step.

To make work on the system easier, we denote the first equation R1, the second R2 and third R3:

x + y + 2z = -2 --> R1

3x - y + 14z = 6 --> R2

x + 3y = -5 --> R3

The equation R3 has coefficient of z equal to 0. We can eliminate the z variable in the second equation by multiplying R1 by 7 and subtract R2:

7R1 --> 7x + 7y + 14z = -14

R2 --> 3x -y + 14z = 6

7R1 - R2 --> 4x + 8y = -20

The system looks like

x + y + 2z = -2 --> R1

4x + 8y = -20 --> R2

x + 3y = -5 --> R3

(This step makes calculations easier) We divide R2 by 4.

R2/4 --> x + 2y = -5.

The system:

x + y + 2z = -2 --> R1

x + 2y = -5 --> R2

x + 3y = -5 --> R3

The last step before we start substituting R2 - R3:

R2 --> x + 2y = -5

R3 --> x + 3y = -5

R2 -R3 --> -y = 0.

We see that y = 0.

Now, we substitute y in R2 or R3 and find:

x + 2x0 = -5

x = - 5 - 0 = -5.

Next, we use x and y and equation R1 to find z:

(-5) + 0 + 2z = -2

2z - 5 = -2,

z = 1.5.

The check of the system is left as an exercise.

I suggest you always check the found solution of a system by substituting all values in the original equations, so you see you did not make a mistake.

Any questions?