Lesson BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method

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BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method

This lesson follows the lesson Solving systems of linear equations in 3 unknowns by the Substitution method in this site.
The referred lesson was quite long. It is because the theme is quite complicated and requires detailed explanations.
But I think that the student will benefit having the short didactic version (instead of and together with the long lesson :) ).

The Substitution method for solving systems of linear equations in 3 unknowns is to express one variable via two others using one equation and then to substitute this expression into the two remaining equations. In this step you reduce the original system of three linear equations in three unknowns to the system of two linear equations in two unknowns.

When this step is done, you solve the obtained system of two linear equations by repeatedly applying the substitution method as described in the lesson Solution of the linear system of two equations in two unknowns by the Substitution method of the section Algebra-I in this site. For it, you express the second unknown via the third one using one of the two linear equations of the obtained system, and then substitute it into the remaining equation of this system.

After completing this, you will get the single linear equation in one unknown which you can easily solve. Then back-substitute the found value into the intermediate system of two linear equations in two unknowns to get the second unknown. As the last step, back-substitute the two found values for unknowns into either appropriate of the original equations - it will allow you to find the last unknown.

Example 1

Example 2

Example 3

It is not necessary to start the substitution method from the very first equation. You can start from any appropriate equation.
It is not necessary to start the substitution method from the very first unknown. You can start from any appropriate unknown.
If there is an equation and an unknown with the coefficient "1" in your system, you may prefer to start the substitution method from this equation and this unknown.

If you are offered to solve the system of linear equations in 3 unknowns with the diagonal coefficient matrix (its profile is shown in the Figure 1), do not use the substitution method. You can reach the solution straight-forward.

Figure 1. The profile of a
diagonal coefficient matrix

Figure 2a. The profile of an
upper triangular coefficient matrix

Figure 2b. "Masked" upper
triangular coefficient matrix

Figure 3a. The profile of an
lower triangular coefficient matrix

Figure 3b. "Masked" lower
triangular coefficient matrix

If you need to solve a system with an upper triangular coefficient matrix (the profile is in the Figure 2a), you should not apply the substitution method in its full range. You should make the back-substitution step only. Same if you are given a system with the masked upper triangular matrix having the profile like in Figure 2b.

Similarly, if you need to solve a system with a lower triangular coefficient matrix (the profile is in the Figure 3a), you should not apply the substitution method in its full range. It is enough to make the straight-forward substitution only. Same if you are given a system with the masked lower triangular matrix having the profile like in Figure 3b.

When solving a systems of three linear equations in three unknowns, you can meet one of three cases:
    1) the case when the system has a unique solution;
    2) the case when the system has infinitely many solutions, and
    3) the case when the system has no solutions.

The substitution method allows you to recognize, to distinct and to identify these cases.

If by applying the substitution method you finally get a single equation of the form a%2Ax+=+b

with a non-zero coefficient

, then it is the first case: the system
has a unique solution, and you can get it in the back-substitution process.

If by applying the substitution method you finally get a single equation of the form 0%2Ax+=+f

with the zero coefficient

(or, generally,
an equation a%5Bi1%5D%2Ax+%2B+a%5Bi2%5D%2Ay+%2B+a%5Bi3%5D%2Az

with zero coefficients a%5Bi1%5D

and

, where

are the modified matrix coefficients and f%5Bi%5D

is the modified right
side), then it is the second or the third case.

Further, if the right side

(or

in the general case) is equal to zero, then it is the second case: the original system has infinitely many solutions, and
you can get them in the back-substitution process.
Otherwise, if the right side

(or

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