### 7.2 - The substitution method

Click here if you want instructions *on using the Algebra Coach*
to carry out the substitution method.

In this section we *explain* the substitution method.
Suppose we have a system of *n* linear equations in *n* variables or unknowns.
In the substitution method, we pick one equation and solve it for one of the unknowns,
say *x*. Then we substitute this solution for *x* into the other
*n* − 1 equations wherever the variable *x* appears, and simplify.
The result is that the *n* − 1 equations contain only
*n* − 1 unknowns (*x* no longer appears).

We repeat this process until we get 1 equation in 1 unknown, which is then easily solved.
The final step is to back-substitute the solution for *x* into the previous
equations to find the values of all the other unknowns.

**Example:** Solve the system:
**Solution:** We arbitrarily choose to solve the first equation for *x*:
*x* = 4 − 1 *y* − 2 *z*.

Substituting this into the second and third equations we get:
Simplifying these equations we get:
Note that we now have 2 equations in 2 unknowns. We repeat the process.
We arbitrarily choose to solve the first of these equations for *y*:
*y* = −1.2 *z* + 0.4

Substituting this expression into the second equation gives:
−3(−1.2 z + 0.4) − 5 *z* = 0.

Simplifying gives:
*z* = −0.8571

Now that we have the solution for *z*, we enter the back-substitute phase
to find the values of *x* and *y*. First substitute the value
for *z* back into either of the previous pair of equations:
Either choice yields *y* = 1.429. Now substitute the values for both
*z* and *y* back into any of the original three equations:
Any choice yields *x* = 4.286. Thus the final solution (to 3 significant figures) is:
{ *x* = 4.29, *y* = 1.43, *z* = −0.857 }.

### Less equations than unknowns

If the number of equations in the system is less than the number of unknowns then
the last time you repeat the above process you will have one equation
remaining but more than one unknown remaining. You will not be able to get a unique value
for any variable; the best you will be able to do is solve for one variable in terms
of the others, and back-substitute that *expression*. Try the exercises,
which contain examples of systems with less equations than unknowns.

### Recognizing Redundant or Inconsistent Systems

If the number of equations is greater than the number of unknowns then the
systems is guaranteed to be either redundant
or inconsistent. But if the number of
equations is equal to or less than the number of unknowns then you will generally
not recognize a system as being redundant or inconsistent
until the very end of the calculation.
This is especially true if the system is large.

If you are solving the system of equations by the substitution method and the
system is redundant, then you will end up with a final equation that states
0 = 0. If the system is inconsistent, then you will end up with one that
states a contradiction like 0 = 5. In either case back-substitution is
impossible so the best you can do is leave the system of equations in its
present state. Try the exercises, which contain examples of
redundant and inconsistent systems of equations.