- (a)
- Is the system linear in form?
- (b)
- If so, make substitutions by defining variables and so that the system in terms of and is linear. What is ? What is ? What is our new associated linear system?
- (c)
- What is the solution set to our associated linear system?
- (d)
- What is the solution set to our original system?
Solving Non-linear Systems Algebraically
Algebraically, we can use the methods of substitution and elimination outlined in Section 8.1 to solve non-linear systems of equations. However, we need to exercise care when solving non-linear systems, especially since the operations involved may not always result in valid solutions!
For example, consider the system given by
Let’s try to use substitution here. From the top equation, we can see that . Substituting this into the bottom equation results in , or , which we can rewrite as . We can now use the quadratic formula on to find that . Taking a square root, we find that are possible values of . Note that there are actually four separate possible values of , one for each choice of plus or minus in the expression above: , and .
However, is actually negative! Since the square root of a negative number is not a real number, and are not valid -values of a solution to this system. The other two solutions are fine. Therefore, keeping in mind that , the solutions to our system are given by and . We can plug these back into the original equations to make sure that they satisfy both.
Taking a look at the graphs of these equations should shed some light on what’s happening here.
The only intersection points of the two graphs have a positive -coordinate. This could have tipped us off earlier that some of the -values we got wouldn’t be valid. Indeed, from the first equation, we have that , and this ensures that must be positive.
The above example illustrates the importance of always checking that the solutions you find are real numbers, and also checking that the solutions you find are actually solutions to the system.
Eliminating Variables
Now we illustrate the method of elimination, which can be used when you notice that the equations in the system have like terms. The difference from before is that we now may have non-linear terms that we can eliminate.
Let’s apply this technique to a system we saw previously.
We can multiply the top equation by , so we get the equivalent system of
equations
Now we can eliminate the terms to obtain . From here, we see that , so . To find the
associated values, we substitute each value of into one of the equations to find the
resulting value of . Choosing , we find that for both and , we get . Our solution set is
thus .
We proceed as before to eliminate one of the variables. We can multiply the top
equation by , so we get the equivalent system of equations
Now we can eliminate the terms to obtain . From here, we see that . Since the
square root of a negative number is not a real number, we see that there are no
real values of that solve this equation. Therefore, we conclude that this
system has no solution. Recall that a system that has no solution is called
inconsistent.
At first glance, it doesnt appear as though elimination will do us any good since its
clear that we cannot completely eliminate one of the variables. The alternative,
solving one of the equations for one variable and substituting it into the
other, is full of unpleasantness. Returning to elimination, we note that it is
possible to eliminate the troublesome term, and the constant term as well, by
elimination and doing so we get a more tractable relationship between and .
We can multiply the top equation by , so we get the equivalent system of
equations
Eliminating, we find that , so , and . Substituting into the top equation, we get , so
that or . On the other hand, when we substitute into the top equation, we get or ,
which gives no real solutions. Substituting each of into the substitution equation
yields the solution set . Try plugging these into the original system to see that they
are actually solutions. Verifying this graphically would be a fun exercise, but we leave
that up to you.
Some Common Issues and Techniques
Notice that this is, in fact, a non-linear system, since the second equation contains
an term. Since we can’t see any like terms in the two equations, it makes sense to try
to use substitution. We might be tempted to divide both sides of the bottom
equation by , so as to isolate , but as always with division, we need to be careful!
Indeed, is still a possibility, so we cannot divide through by , since we’d then be
dividing by .
Instead, it helps to think about what it means for the product of two numbers to equal 0. In fact, the product of two nonzero numbers can never be 0. In our situation, we know that , so either or .
If , then dividing by 3 (since ) gives us . We can plug that into the top equation and find that , so . We can then check that is a solution to our original system.
If , we can plug that into the top equation to find that . Solving for yields . We can then check that and are solutions to the system.
Our final solution set is .
Notice that this is, in fact, a non-linear system, since both equations divide by the
variables we’re using.
If we define new (but related) variables by letting and then the system becomes
This associated system of linear equations can then be solved using any of the techniques you’ve learned earlier to find that and . Therefore, and , and our solution set is .
We say that the original system is linear in form because its equations are not linear, but a few substitutions reveal a structure that we can treat like a system of linear equations. However, the substitutions may introduce some complexity, as seen in the following example.
Notice that this is, in fact, a non-linear system, since both equations divide by the
variables we’re using.
If we define new (but related) variables by letting and then the system becomes
This associated system of linear equations can then be solved using any of the techniques you’ve learned earlier to find that and . Therefore, and . However, the nonlinearity of the system throws us a wrench! Logarithms are not defined on negative numbers, so does not exist, and there is actually no value of that satisfies both equations. Therefore, this system does not have a solution.
Consider the following system.
- When solving non-linear systems, it’s important to check that your solutions exist.
- Substitution and elimination can be used when solving non-linear systems.
- Some non-linear systems are linear in form, meaning that substituting new variables yields a linear system that can be solved. Take care when undoing the substitutions: not everything may be defined!