A linear system of two equations with two variables is any
system that can be written in the form.
where any of the constants can be zero with the exception
that each equation must have at least one variable in it.
Also, the system is called linear if the variables are only
to the first power, are only in the numerator and there are no products of
variables in any of the equations.
Here is an example of a system with numbers.
Before we discuss how to solve systems we should first talk
about just what a solution to a system of equations is. A solution to a system of equations is a
value of x and a value of y that, when substituted into the
equations, satisfies both equations at the same time.
For the example above and is a solution to the system. This is easy enough to check.
So, sure enough that pair of numbers is a solution to the system. Do not worry about how we got these
values. This will be the very first
system that we solve when we get into examples.
Note that it is important that the pair of numbers satisfy
both equations. For instance and will satisfy the first equation, but not the
second and so isn’t a solution to the system.
Likewise, and will satisfy the second equation but not the
first and so can’t be a solution to the system.
Now, just what does a solution to a system of two equations
represent? Well if you think about it
both of the equations in the system are lines.
So, let’s graph them and see what we get.
As you can see the solution to the system is the coordinates
of the point where the two lines intersect.
So, when solving linear systems with two variables we are really asking
where the two lines will intersect.
We will be looking at two methods for solving systems in
The first method is called the method of substitution. In
this method we will solve one of the equations for one of the variables and
substitute this into the other equation.
This will yield one equation with one variable that we can solve. Once this is solved we substitute this value
back into one of the equations to find the value of the remaining variable.
In words this method is not always very clear. Let’s work a couple of examples to see how
this method works.
Example 1 Solve
each of the following systems.
So, this was the first system that we looked at
above. We already know the solution,
but this will give us a chance to verify the values that we wrote down for
Now, the method says that we need to solve one of the
equations for one of the variables.
Which equation we choose and which variable that we choose is up to
you, but it’s usually best to pick an equation and variable that will be easy
to deal with. This means we should try
to avoid fractions if at all possible.
In this case it looks like it will be really easy to solve
the first equation for y so let’s
Now, substitute this into the second equation.
This is an equation in x
that we can solve so let’s do that.
So, there is the x
portion of the solution.
Finally, do NOT forget to go back and find the y portion of the solution. This is one of the more common mistakes
students make in solving systems. To
so this we can either plug the x
value into one of the original equations and solve for y or we can just plug it into our substitution that we found in
the first step. That will be easier so
let’s do that.
So, the solution is and as we noted above.
[Return to Problems]
With this system we aren’t going to be able to completely
avoid fractions. However, it looks
like if we solve the second equation for x
we can minimize them. Here is that
Now, substitute this into the first equation and solve the
resulting equation for y.
Finally, substitute this into the original substitution to
So, the solution to this system is and .
[Return to Problems]
As with single equations we could always go back and check
this solution by plugging it into both equations and making sure that it does
satisfy both equations. Note as well
that we really would need to plug into both equations. It is quite possible that a mistake could
result in a pair of numbers that would satisfy one of the equations but not the
Let’s now move into the next method for solving systems of
equations. As we saw in the last part of
the previous example the method of substitution will often force us to deal
with fractions, which adds to the likelihood of mistakes. This second method will not have this
problem. Well, that’s not completely
true. If fractions are going to show up
they will only show up in the final step and they will only show up if the
solution contains fractions.
This second method is called the method of elimination. In
this method we multiply one or both of the equations by appropriate numbers (i.e. multiply every term in the equation
by the number) so that one of the variables will have the same coefficient with
opposite signs. Then next step is to add
the two equations together. Because one
of the variables had the same coefficient with opposite signs it will be
eliminated when we add the two equations.
The result will be a single equation that we can solve for one of the
variables. Once this is done substitute
this answer back into one of the original equations.
As with the first method it’s much easier to see what’s
going on here with a couple of examples.
Example 2 Solve
each of the following systems of equations.
This is the system in the previous set of examples that
made us work with fractions. Working
it here will show the differences between the two methods and it will also
show that either method can be used to get the solution to a system.
So, we need to multiply one or both equations by constants
so that one of the variables has the same coefficient with opposite
signs. So, since the y terms already have opposite signs
let’s work with these terms. It looks
like if we multiply the first equation by 3 and the second equation by 2 the y terms will have coefficients of 12
and -12 which is what we need for this method.
Here is the work for this step.
So, as the description of the method promised we have an
equation that can be solved for x. Doing this gives, which is exactly what we found in the
previous example. Notice however, that
the only fraction that we had to deal with to this point is the answer itself
which is different from the method of substitution.
Now, again don’t forget to find y. In this case it will be
a little more work than the method of substitution. To find y
we need to substitute the value of x
into either of the original equations and solve for y. Since x is a fraction let’s notice that, in
this case, if we plug this value into the second equation we will lose the
fractions at least temporarily. Note
that often this won’t happen and we’ll be forced to deal with fractions
whether we want to or not.
Again, this is the same value we found in the previous example.
[Return to Problems]
In this part all the variables are positive so we’re going
to have to force an opposite sign by multiplying by a negative number
somewhere. Let’s also notice that in
this case if we just multiply the first equation by -3 then the coefficients
of the x will be -6 and 6.
Sometimes we only need to multiply one of the equations
and can leave the other one alone.
Here is this work for this part.
Finally, plug this into either of the equations and solve
for x. We will use the first equation this time.
So, the solution to this system is and .
[Return to Problems]
There is a third method that we’ll be looking at to solve
systems of two equations, but it’s a little more complicated and is probably
more useful for systems with at least three equations so we’ll look at it in a
Before leaving this section we should address a couple of
special case in solving systems.
Example 3 Solve
the following systems of equations.
We can use either method here, but it looks like
substitution would probably be slightly easier. We’ll solve the first equation for x and substitute that into the second
So, this is clearly not true and there doesn’t appear to
be a mistake anywhere in our work. So,
what’s the problem? To see let’s graph
these two lines and see what we get.
It appears that these two lines are parallel (can you
verify that with the slopes?) and we know that two parallel lines with
different y-intercepts (that’s
important) will never cross.
As we saw in the opening discussion of this section
solutions represent the point where two lines intersect. If two lines don’t intersect we can’t have
So, when we get this kind of nonsensical answer from our
work we have two parallel lines and there is no solution to this system of equations.
The system in the previous example is called inconsistent. Note as well that if we’d used elimination on
this system we would have ended up with a similar nonsensical answer.
Example 4 Solve
the following system of equations.
In this example it looks like elimination would be the
On first glance this might appear to be the same problem
as the previous example. However, in
that case we ended up with an equality that simply wasn’t true. In this case we have 0=0 and that is a true
equality and so in that sense there is nothing wrong with this.
However, this is clearly not what we were expecting for an
answer here and so we need to determine just what is going on.
We’ll leave it to you to verify this, but if you find the
slope and y-intercepts for these
two lines you will find that both lines have exactly the same slope and both
lines have exactly the same y-intercept. So, what does this mean for us? Well if two lines have the same slope and
the same y-intercept then the
graphs of the two lines are the same graph.
In other words, the graphs of these two lines are the same graph. In these cases any set of points that
satisfies one of the equations will also satisfy the other equation.
Also, recall that the graph of an equation is nothing more
than the set of all points that satisfies the equation. In other words, there is an infinite set of
points that will satisfy this set of equations.
In these cases we do want to write down something for a
solution. So what we’ll do is solve
one of the equations for one of the variables (it doesn’t matter which you
choose). We’ll solve the first for y.
Then, given any x
we can find a y and these two
numbers will form a solution to the system of equations. We usually denote this by writing the
solution as follows,
So show that these give solutions let’s work through a
couple of values of t.
To show that this is a solution we need to plug it into
both equations in the system.
So, and is a solution to the system. Let’s do another one real quick.
Again we need to plug it into both equations in the system
to show that it’s a solution.
Sure enough and is a solution.
So, since there are an infinite number of possible t’s there must be an infinite number of solutions to this
system and they are given by,
Systems such as those in the previous examples are called dependent.
We’ve now seen all three possibilities for the solution to a
system of equations. A system of
equation will have either no solution, exactly one solution or infinitely many