College Algebra
Tutorial 49: Solving Systems of
Linear Equations in Two Variables
 Learning Objectives
After completing this tutorial, you should be able to:
- Know if an ordered pair is a solution to a system of
linear equations
in
two variables or not.
- Solve a system of linear equations in two variables
by graphing.
- Solve a system of linear equations in two variables
by the substitution
method.
- Solve a system of linear equations in two variables
by the elimination
by addition method.
|
Introduction
In this tutorial we will be specifically looking at
systems that have
two equations and two unknowns. Tutorial 50: Solving Systems of
Linear
Equations in Three Variables will cover systems that have three
equations
and three unknowns. We will look at solving them three different
ways:
by graphing, by the substitution method, and by the elimination by
addition
method.
So, let's go ahead and look at these systems. |
Tutorial
|
System of Linear Equations
|
A system of linear equations is two or more linear
equations that
are being solved simultaneously.
In this tutorial, we will be looking at systems that
have only two linear
equations and two unknowns. |
In general, a solution of a system in two variables
is an ordered
pair that makes BOTH equations true.
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.
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 line.
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. |
There are three possible
outcomes that you
may encounter when working with these systems:
-
-
-
|
One Solution
If the system in two variables has one solution, it is
an ordered
pair that is a solution to BOTH equations. In other words,
when
you plug in the values of the ordered pair it makes BOTH equations
TRUE.
If you do get one solution 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 one 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 one solution:
|
No Solution
If the two lines are parallel to each other, they will
never intersect.
This means they do not have any points in common. In this
situation,
you would have no solution.
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:
|
Infinite
Solutions
If the two lines end up lying on top of each other, then
there is
an infinite number of solutions. In this situation, they
would
end up being the same line, so any solution that would work in one
equation
is going to work in the other.
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:
|
 Example
1: Determine whether each ordered pair is a solution
of
the system.
(3, 1) and (0, -1) |
Let’s check the ordered pair (3, 1) in the first
equation: |
|
*Plug in 3 for x and 1 for y
*True statement |
So far so good, (3, 1) is a solution to the first
equation 2x - 3y = 3.
Now, let’s check (3, 1) in the second equation: |
|
*Plug in 3 for x and 1 for y
*True statement |
Hey, we ended up with another true statement, which
means (3, 1) is
also a solution to the second equation 4x -
2y = 10.
Here is the big question, is (3, 1) a solution to the
given system?????
Since it was a solution to BOTH equations in the
system, then it
is a solution to the overall system.
Now let’s put (0, -1) into the first equation: |
|
*Plug in 0 for x and -1 for y
*True statement |
This is a true statement, so (0, -1) is a solution to
the first equation
2x - 3y =
3.
Finally, let’s put (0, -1) into the second equation: |
|
*Plug in 0 for x and -1 for y
*False statement |
This time we got a false statement, you know what that
means.
(0, -1) is NOT a solution to the second equation 4x - 2y = 10.
Here is the big question, is (0, -1) a solution to
the given system?????
Since it was not a solution to BOTH equations in the
system, then
it is not a solution to the overall system. |
Three Ways to
Solve Systems of Linear
Equations in Two Variables
|
Step
1: Graph the first equation. |
Unless the directions tell you differently, you can use
any "legitimate"
way to graph the line. If you need a review on graphing lines,
feel free to go back to Tutorial 27: Graphing Lines.
|
Step
2: Graph the second equation on
the same coordinate
system as the first. |
You graph the second equation the same as any other
equation.
Refer to the first step if you need to review how to
graph
a line. The difference here is you will put it on the same
coordinate system
as the first. It is like having two graphing problems in one. |
Step
3: Find the solution. |
If the two lines intersect at one place, then
the point of
intersection is the solution to the system.
If the two lines are parallel, then they never
intersect, so
there is no solution.
If the two lines lie on top of each other, then
they are the
same line and you have an infinite number of solutions.
In this case you can write down either equation as the solution to
indicate
they are the same line. |
Step
4: Check the proposed ordered
pair solution in
BOTH equations. |
You can plug in the proposed solution into BOTH
equations. If
it makes BOTH equations true then you have your solution to the
system.
If it makes at least one of them false, you need to go
back and redo
the problem. |
Example
2: Solve the system of equation by graphing.
 |
|
*Plug in 0 for y for x-int
*x-intercept
|
The x-intercept is (3, 0).
y-intercept |
|
*Plug in 0 for x for y-int
*y-intercept
|
The y-intercept is (0, 3).
Find another
solution by letting x = 1. |
Another solution is (1, 2).
Solutions:
|
x
|
y
|
(x, y)
|
|
3
|
0
|
(3, 0)
|
|
0
|
3
|
(0, 3)
|
|
1
|
2
|
(1, 2)
|
Plotting the ordered pair solutions and drawing the
line:
|
|
*Plug in 0 for y for x-int
*x-intercept
|
The x-intercept is (1, 0).
y-intercept |
|
*Plug in 0 for x for y-int
*Inverse of mult. by -1 is div.
by -1
*y-intercept |
The y-intercept is (0, -1).
Find another
solution by letting x = 2. |
|
*Plug in 2 for x
*Inverse of add 2 is sub. 2
*Inverse of mult. by -1 is div
by -1
|
Another solution is (2, 1).
Solutions:
|
x
|
y
|
(x, y)
|
|
1
|
0
|
(1, 0)
|
|
0
|
-1
|
(0, -1)
|
|
2
|
1
|
(2, 1)
|
Plotting the ordered pair solutions and drawing the
line:
|
We need to ask ourselves, is there any place that the
two lines intersect,
and if so, where?
The answer is yes, they intersect at (2, 1). |
You will find that if you plug the ordered pair (2, 1)
into BOTH equations
of the original system, that this is a solution to BOTH of them.
The solution to this system is (2, 1). |
Example
3: Solve the system of equation by graphing.
 |
|
*Plug in 0 for y for x-int
*x-intercept
|
The x-intercept is (5, 0).
y-intercept |
|
*Plug in 0 for x for y-int
*y-intercept |
The y-intercept is (0, 5).
Find another
solution by letting x = 1. |
|
*Plug in 1 for x
*Inverse of add 1 is sub. 1
|
Another solution is (1, 4).
Solutions:
|
x
|
y
|
(x, y)
|
|
5
|
0
|
(5, 0)
|
|
0
|
5
|
(0, 5)
|
|
1
|
4
|
(1, 4)
|
Plotting the ordered pair solutions and drawing the
line:
|
|
*Plug in 0 for y for x-int
*Inverse of add 3 is sub. 3
*Inverse of mult. by -1 is div.
by -1
*x-intercept |
The x-intercept is (3, 0).
y-intercept |
|
*Plug in 0 for x for y-int
*y-intercept
|
The y-intercept is (0, 3).
Find another
solution by letting x = 1. |
Another solution is (1, 2).
Solutions:
|
x
|
y
|
(x, y)
|
|
3
|
0
|
(3, 0)
|
|
0
|
3
|
(0, 3)
|
|
1
|
2
|
(1, 2)
|
Plotting the ordered pair solutions and drawing the
line:
|
We need to ask ourselves, is there any place that the
two lines intersect,
and if so, where?
The answer is no, they do not intersect. We
have two parallel
lines. |
There are no ordered pairs to check.
The answer is no solution. |
|
Solve by the Substitution Method
|
Step 1: Simplify if needed. |
This would involve things like removing ( ) and
removing fractions.
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. |
Step 2: Solve one
equation for either variable. |
It doesn't matter which equation you use or which
variable you choose
to solve for.
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 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!!). |
Step 3: Substitute what
you get for step 2 into the other equation. |
This is why it is called the substitution method.
Make sure that
you substitute the expression into the OTHER equation, the one you
didn't
use in step 2. This will give you one equation with one unknown. |
Step 4: Solve for
the remaining variable . |
Solve the equation set up in step 3 for the variable
that is left.
If you need a review on solving linear equations, feel
free to go to Tutorial
14: Linear Equations in On Variable.
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. |
Step 5: Solve for
second variable. |
If you come up with a value for the variable in step
4, that means
the two equations have one solution. Plug the value found in
step 4 into any of the equations in the problem and solve for the other
variable. |
Step 6: Check the proposed
ordered pair solution in BOTH original equations. |
You can plug in the proposed solution into BOTH
equations. If
it makes BOTH equations true, then you have your solution to the
system.
If it makes at least one of them false, you need to go
back and redo
the problem. |
Example 4: Solve the system of equations
by the substitution
method:
 |
Both of these equations are already simplified.
No work needs
to be done here. |
It does not matter which equation or which variable you
choose to solve
for. Just keep it simple.
Since the x in the second
equation has a
coefficient of 1, that would mean we would not have to divide by a
number
to solve for it and run the risk of having to work with fractions
(YUCK).
The easiest route here is to solve the second equation for x,
and we definitely want to take the easy route.
You would not be wrong to either choose the other
equation and/or solve
for y, again you want to keep it as
simple
as possible.
Solving the second equation for x we
get: |
|
*2nd equation solved for x |
Substitute the expression y + 1 for x into the first equation and solve for y:
(when you plug in an expression like this, it is just like you plug
in a number for your variable) |
|
*Sub. y + 1 in
for x
*Dist. 3 through ( )
*Combine like terms
*Inverse of add 3 is sub. 3
|
Plug in 3 for y into
the equation in
step 2 to find x’s value. |
You will find that if you plug the ordered pair (4, 3)
into BOTH equations
of the original system, that this is a solution to BOTH of them.
(4, 3) is a solution to our system. |
Example 5: Solve the system of equations
by the substitution
method:
 |
This equation is full of those nasty fractions.
We can simplify
both equations by multiplying each separate one by it’s LCD, just like
you can do when you are working with one equation. As long as you
do the same thing to both sides of an equation, you keep the two sides
equal to each other. Multiplying each equation by it's respective LCD we
get: |
|
*Mult. by LCD of 2
*Mult. by LCD of 2
|
Note how the second equation is already solved for y.
We can use that one for this step.
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: |
Substitute the expression -3x +
4 for y into the first equation and solve for x:
(when you plug in an expression like this, it is just like you plug
in a number for your variable) |
|
*Sub. -3x + 4
for y
*Variable dropped out AND false
|
Wait a minute, where did our
variable go????
As mentioned above if your variable drops out and you
have a FALSE statement,
then there is no solution. If we were to graph these two,
they would be parallel to each other. |
Since we did not get a value for x,
there
is nothing to plug in here. |
There are no ordered pairs to check.
The answer is no solution. |
Example 6: Solve the system of equations
by the substitution
method:
 |
Both of these equations are already simplified.
No work needs
to be done here. |
Note how the second equation is already solved for y.
We can use that one for this step.
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: |
Substitute the expression 2x -
4 for y into the first equation and solve for x:
(when you plug in an expression like this, it is just like you plug
in a number for your variable) |
|
*Sub. 2x - 4 for y
*Variable dropped out AND true |
Wait a minute, where did our
variable go????
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 line. |
Since we did not get a value for x,
there
is nothing to plug in here. |
There is no value to plug in here.
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 {(x, y)| 2x - y = 4} OR {(x, y)
| y = 2x -
4}. |
|
Solve by the Elimination by
Addition Method
|
Step 1: Simplify and
put both equations in the form Ax + By = C if needed. |
This would involve things like removing ( ) and
removing fractions.
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. |
Step 2: Multiply one
or both equations by a number that will create opposite coefficients
for
either x or y if needed. |
Looking ahead, we will be adding these two
equations together.
In that process, we need to make sure that one of the variables drops
out,
leaving us with one equation and one unknown. The only way we can
guarantee that is if we are adding opposites. The sum of
opposites
is 0.
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. |
Add the two equations together.
The variable that has the opposite coefficients will
drop out in this
step and you will be left with one equation with one unknown. |
Step 4: Solve for remaining
variable. |
Solve the equation found in step 3 for the variable
that is left.
If you need a review on solving linear equations, feel
free to go to Tutorial
14: Linear Equations in On Variable.
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. |
Step 5: Solve
for second
variable. |
If you come up with a value for the variable in step
4, that means
the two equations have one solution. Plug the value found in
step 4 into any of the equations in the problem and solve for the other
variable. |
Step 6: Check
the proposed
ordered pair solution in BOTH original equations. |
You can plug the proposed solution into BOTH
equations. If it
makes BOTH equations true, then you have your solution to the
system.
If it makes at least one of them false, you need to go
back and redo
the problem. |
Example 7: Solve the system of equation by
the elimination
method:
 |
This equation is full of those nasty fractions.
We can simplify
both equations by multiplying each separate one by it’s LCD, just like
you can do when you are working with one equation. As long as you
do the same thing to both sides of an equation, you keep the two sides
equal to each other. Multiplying each equation by it's respective LCD we
get: |
|
*Mult. by LCD of 6
*Mult. by LCD of 40
|
Again, you want to make this as simple as
possible.
Note how the coefficient on y in the first
equation is 2 and in the second equation it is 5. We need to have
opposites, so if one of them is 10 and the other is -10, they would
cancel
each other out when we go to add them. If we added them together
the way they are now, we would end up with one equation and two
variables,
nothing would drop out. And we would not be able to solve it.
So I proposed that we multiply the first equation by
5 and the second
equation by -2, this would create a 10 and a -10 in front of the y’s
and we will have our opposites.
Multiplying the first equation by 5 and the second
equation by -2
we get: |
|
*Mult. both sides of 1st eq.
by 5
*Mult. both sides of 2nd eq. by -2
*y's
have opposite
coefficients
|
|
*Note that y's
dropped out
|
You can choose any equation used in this problem to
plug in the found x value.
I choose to plug in 10 for x into the
first simplified equation (found in step 1) to find y’s
value. |
|
*Plug in 10 for x
*Inverse of add 30 is sub. 30
*Inverse of mult. by 2 is div.
by 2
|
You will find that if you plug the ordered pair (10,
24) into BOTH
equations of the original system, that this is a solution to BOTH of
them.
(10, 24) is a solution to our system. |
Example 8: Solve the system of equation by
the elimination
method:
 |
This problem is already simplified, however the second
equation is
not written in the form Ax + By = C. In other words, we need to write it in this form so
everything
is lined up ready to go when we add the two equations together. Rewriting the second equation we get: |
|
*Inverse of add x is sub. x
*Everything is lined up
|
Note that x’s
coefficients are already
opposites. y’s coefficients are
also
opposites for that matter. So we do not have to multiply either equation by a
number. |
|
*Note that x's
and y's both dropped out
|
Hey where did our variables
go??
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 line. |
There is no value to plug in here. |
There is no value to plug in here.
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 {(x, y)| x - y = 3} OR {(x, y)
| y = x -
3}. |
Example 9: Solve the system of equation by
the elimination
method:
 |
Both of these equations are already simplified and in
the right form.
No work needs to be done here. |
Again, you want to make this as simple as
possible.
Note how the coefficient on x in the first
equation is 5 and in the second equation it is 10. We need to
have
opposites, so if one of them is -10 and the other is 10, they would
cancel
each other out when we go to add them. If we added them together
the way they are now, we would end up with one equation and two
variables,
nothing would drop out. And we would not be able to solve it.
So I proposed that we multiply the first equation by
-2, this
would create a -10 and a 10 in front of the x’s
and we will have our opposites.
Multiplying the first equation by -2 we get: |
|
*Mult. both sides of 1st eq.
by -2
*x's
and y's
have opposite coefficients
|
|
*Note that x's
and y's both dropped out
|
Wait a minute, where did our
variable go????
As mentioned above if your variable drops out and you
have a FALSE statement,
then there is no solution. If we were to graph these two,
they would be parallel to each other. |
There is no value to plug in here. |
There is no value to plug in here.
The answer is no solution. |
Practice Problems
These are practice problems to help bring you to the next level.
It will allow you to check and see if you have an understanding of these
types of problems. Math works just like anything
else, if you want to get good at it, then you need to practice it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.
In fact there is no such thing as too much practice.
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. |
 Practice Problems 1a - 1c: Solve each system by either the
substitution
or elimination by addition method.
Practice Problem 2a: Solve the system by graphing.
Need Extra Help on these Topics?
Last revised on March 25, 2011 by Kim Seward.
All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.
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