Note that in a nonlinear system, one of your equations can be linear, just not all of them.

In this tutorial, we will be looking at systems that have only two equations and two unknowns. 

In other words, it is where the two graphs intersect, what they have in common.  So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system.

Since we are looking at nonlinear systems, in some cases, there may be more than one ordered pair that satisfies all equations in the system.

A consistent system is a system that has at least one solution.

An inconsistent system is a system that has no solution.
 

The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other equation.  In other words, they end up being the same graph.

The equations of a system are independent if they do not share ALL solutions.  They can have one point in common, just not all of them.

1. a finite number of solutions
2. no solution
3. infinite solutions

Since we are looking at nonlinear systems, in some cases there may be more than one ordered pair that satisfies all equations in the system.

If you do get a finite number of solutions for your final answer, is this system consistent or inconsistent?
If you said consistent, give yourself a pat on the back! 

If you do get a finite number of solutions for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
 

The graph below illustrates a system of two equations and two unknowns that has four solutions:

If you get no solution for your final answer, is this system consistent or inconsistent?
If you said inconsistent, you are right! 

If you get no solution for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
 

The graph below illustrates a system of two equations and two unknowns that has no solution:

If you get an infinite number of  solutions for your final answer, is this system consistent or inconsistent?
If you said consistent, you are right! 

If you get an infinite number of  solutions for your final answer, would the equations be dependent or independent?
If you said dependent, you are correct!
 

The graph below illustrates a system of two equations and two unknowns that has an infinite number of solutions:

To remove ( ): just use the distributive property. 

To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of  your  fractions.

You want to make it as simple as possible.  If one of the equations is already solved for one of the variables, that is a quick and easy way to go.

If you need to solve for a variable, then try to pick one that has a 1 or -1 as a coefficient.  That way when you go to solve for it, you won't have to divide by a number and run the risk of having to work with a fraction (yuck!!).

Also, it is easier to solve for a variable that is to the 1 power, as opposed to being squared, cubed, etc.

This will give you one equation with one unknown. 

Most of the equations in this step will end up being either linear or quadratic.  Once in awhile you will run into a different type of equation.

If you need a review on solving linear or quadratic equations, feel free to go to Tutorial 14: Linear Equations in On Variable or Tutorial 17: Quadratic Equations.

Keep in mind that when you go to solve for this variable that you may end up with no solution for your answer.  For example, you may end up with your variable equaling the square root of a negative number, which is not a real number, which means there would be no solution. 

If your variable drops out and you have a FALSE statement, that means your answer is no solution. 

If your variable drops out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.

If it makes at least one of them false, you need to go back and redo the problem.

Since the y in the first equation has a coefficient of -1, that would mean we would not run the risk of having to work with fractions (YUCK).  The easiest route here is to solve the first equation for y, and we definitely want to take the easy route. 

You would not be wrong to either choose the other equation and/or solve for x, again you want to keep it as simple as possible.

Solving the first equation for y we get:

*1st equation solved for y

*Set 1st factor = 0
 
 
 
 
 

*Set 2nd factor = 0

Plug in 2 for x into the equation in step 2 to find y's value.

(-1/2, 1/2) and (2, 8) are both a solution to our system.

It does not matter which equation or which variable you choose to solve for.  But it is to your advantage to keep it as simple as possible. 

Second equation solved for y:

*Square both binomials to clear the (   )
 
 
 
 
 

*Solve using the Square Root Method
 
 

*Solved for x
 

As mentioned above, in some cases we end up with no solution.  Here is an example where even though we came up with two values for x, we did not come up with a real number answer for a variable.  This means there is no solution.

The answer is no solution.

To remove ( ): just use the distributive property. 

To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of  your  fractions.

If neither variable drops out, then we are stuck with an equation with two unknowns which is unsolvable.

It doesn't matter which variable you choose to drop out.  You want to keep it as simple as possible.  If a variable already has opposite coefficients than go right to adding the two equations together.  If they don't, you need to multiply one or both equations by a number that will create opposite coefficients in one of your variables.  You can think of it like a LCD.  Think about what number the original coefficients both go into and multiply each separate equation accordingly.  Make sure that one variable is positive and the other is negative before you add. 

For example, if you had a 2x in one equation and a 3x in another equation, we could multiply the first equation by 3 and get 6x and the second equation by -2 to get a -6x.  So when you go to add these two together they will drop out.

Since we are working with nonlinear systems here, you need to keep in mind that the variable that you want to eliminate must be a like variable.  In other words, it must not only be the same variable, but have the same exponent.  For example, you can not combine 2x with -2x squared, they are not like terms. But you can combine 2x squared with -2x squared or 2x with -2x.  This will play a big part on your decision  as to what to eliminate. 

So even though in general, it doesn't matter what variable you choose to drop out, sometimes the problem will dictate that decision for you, because you are unable to eliminate one of the variables because they are not like terms.

The variable that has the opposite coefficients will drop out in this step and you will be left with one equation with one unknown.

Most of the equations in this step will end up being either linear or quadratic.  Once in awhile you will run into a different type of equation.

If you need a review on solving linear or quadratic equations, feel free to go to Tutorial 14: Linear Equations in On Variable or Tutorial 17: Quadratic Equations.

Keep in mind that when you go to solve for this variable that you may end up with no solution for your answer.  For example, you may end up with your variable equaling the square root of a negative number, which is not a real number, which means there would be no solution.

If both variables drop out and you have a FALSE statement, that means your answer is no solution. 

If both variables drop out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.

If it makes at least one of them false, you need to go back and redo the problem.

The variable that you want to eliminate must be a like variable.  In other words, it must not only be the same variable, but have the same exponent.  Note how x is squared on both equations.  So we would be able to get opposite coefficients on them and then when we added the two equations together they would drop out. 

Note how y is to the one power in the first equation and squared on the second equation.  If we would pick y to eliminate we would have a problem because we cannot combine unlike terms together.  y squared -y would be y squared - y NOT 0.

So I proposed that we multiply the second equation by -1, this would create a 1 and a -1 in front of the x squareds and we will have our opposites.
 

Multiplying the second equation by -1 we get:

*x squareds have opposite coefficients
 

*Set first factor = 0
 
 
 
 

*Set second factor = 0

I choose to plug in -4 for y into the first equation to find x's value.

*Solve using the Square Root Method
 

You will find that if you plug the ordered pairs (0, - 4), , and into BOTH equations of the original system, that they would all be solutions to BOTH of them.

(0, - 4), , and are all solutions to our system.

Multiplying the first equation by its LCD we get:

The variable that you want to eliminate must be a like variable. Note that we could either choose to eliminate the x to the fourth terms or the y to the fourth terms or even the x to the fourth y to the fourth terms.  I'm going to choose to eliminate the x to the fourth terms, which means I want opposite coefficient on those terms.  Looks like we just need to multiply the first equation by -2.

Multiplying the first equation by -2 we get:

*x to the fourths have opposite coefficients
 

As mentioned above, if the variable drops out AND we have a TRUE statement, then when have an infinite number of solutions.  They end up being the same graph.

When they end up being the same equation, you have an infinite number of solutions.  You can write up your answer by writing out either equation to indicate that they are the same equation.

Two ways to write the answer are  OR .

Note that when solving these nonlinear systems, you may come across some that you are unable to use the substitution method  and some that you are unable to use the elimination by addition method.  Just try what you feel may work and if it doesn't work try the other method.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that  problem.  At the link you will find the answer as well as any steps that went into finding that answer.