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A system of equations is a combination of multiple algebraic equations that are dependent on one another. You may have seen a system of equations in the form of two algebraic equations with two variables each, $x$ and $y$. This is what we will focus on in this chapter.

Solving a system of equations involves a method that many students find difficult. It is best to be sure that the question you are solving is really a true system of equations before attempting it!

There are two things you should look out for in order to identify a system of equations. First, it will be a question involving two **different** equations. Second, both of the equations will share the same two variables. Examples of problems involving systems of equations are as follows:

Solve the following system of equations for $x$ and $y$.

$x+y=2x=4−y$

Here’s another example:

What is the sum of the solutions to the following system of equations? (The solutions refer to the values of $x$ and $y$.)

$−3y12 =x+4=3x−2y $

One idea to solve a system of equations is solving for one variable at a time, using one equation at a time. This is called the **substitution method**. So, let’s review the steps to solving these kinds of problems.

In the steps below, we will be referencing the following example system of equations:

$x+yy−3 =7=3x $

Since we start with two equations, we will pick the first one to work with. Using this first equation, solve for one of the variables using the basic order of operations. Let’s perform this step, solving for $x$ in our example:

$x+yx =7=7−y $

We found an expression equal to $x$ in the previous step. Now, we will replace $x$ in the second equation with that expression. The result of this is that we will end up with just one variable in one equation. Performing this step on our example, we get:

$y−3y−3 =3x=3∗(7−y) $

Solving for $y$ here will get us an exact value for the variable. That will be the first solution to our system of equations. Let’s go ahead and do that with our example:

$y−3y−34yy =3∗(7−y)=21−3y=24=6 $

Now that we have the exact value of $y$, we can solve for $x$ by plugging that number into either of the original equations for $y$. Below, we will show that both equations give the same solution for $x$:

First equation:

$y−36−3x =3x=3x=1 $

Second equation:

$x+yx+6x =7=7=1 $

Our final solutions to the system of equations are:

$x=1y=6$

This method involves combining both equations into one in order to eliminate one variable. The idea of what we want to do is to manipulate the equations so that both equations have the same value of one variable but with opposite signs. That way, the negative and positive values eliminate the variable from the equations when the equations are added together.

You want to pick a variable in the first equation that can easily be matched by the second equation. For instance, in our example, we would want to pick the variable $y$ because there is the same value of $y$ in the second equation as in the first equation. Let’s rearrange our example equations to make more sense of this: Original equations:

$x+yy−3 =7=3x $

Rearranged equations:

$x+y−3x+y =7=3 $

Looking at our equations this way, we can see that there is a positive $y$ in both equations.

We need to get the $y$ in the second equation to be negative so that when we add the equations together the $y$s will equal out to $0$. In our example, we want to multiply the whole second equation by negative $1$:

Rearranged equations:

$x+y−3x+y =7=3 $

Modified equations:

$x+y3x−y =7=−3 $

Now, we can see that the $y$ values will equal zero when the equations are added together. This means we are ready to proceed to the next step.

We will add the two equations together and combine like terms. Doing this will reduce the equations down to a single equation and the variables down to a single variable. We can then solve this equation for the variable left over after elimination:

$(x+y+3x−y)4xx =(7−3)=4=1 $

We have now found the exact value of $x$ and we are ready to find the value of $y$.

Since we have the value for $x$, we can pick either of the original equations and substitute the calculated value for $x$ to find the value for $y$:

$x+y1+yy =7=7=6 $

So, our final solutions to the system equations are:

$x=1y=6$

Sometimes a system of equations gives abnormal solutions. There could be infinite solutions, or there could be no real solutions. However, there is a way that we can recognize both of these situations.

A system of equations has an infinite number of solutions when the two equations are the same. This also means that if one equation is the same equation as the other but multiplied by a factor, the system of equations has infinite solutions. Below are a few examples of when systems of equations have infinite solutions:

Example:

$x+y=3y+x=3$

Another example:

$y−x2y−2x =2=4 $

We can usually determine if a system of equations has infinite solutions before solving it using either of our methods. When we solve these special systems, we will get a true statement. Let’s solve the first example using the substitution method. Let’s solve for $x$ first:

$x+yx(3−y)+y3 =3=3−y=3=3 $

Three does in fact equal three. It’s always true regardless of the values of $x$ and $y$. This is the solution we get when solving a system of equations that have infinite solutions.

A system of equations has no real solutions when the simplification of the equations leads to a false statement such as $3=7$. Sometimes, this requires us to solve through our system of equations as normal and discover at the end that our numbers do not work out properly. Other times, we may be able to tell by attempting the elimination method and having both variables eliminated at once. The following is an example of a system of equations with no real solutions:

$x+2y4x+8y =2=3 $

Performing our elimination method, we would want to multiply the top equation by $−4$. Adding these modified equations together, we get the following simplification:

$(−4x−8y+4x−8y)0 =(−8+3)=−5 $

This statement is false. It will always be false regardless of the values of $x$ and $y$. Therefore, there are no real solutions to this system of equations.

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