College Algebra:
Solving Linear Equation in two Variables
by Malen dO. Peňalosa BSED 2-A
Partido State University
College of Education
Goa, Camarines Sur
penalosamalen.gmail.com
penalosamalen.yahoomail.com
penalosamalen.gmail.com
penalosamalen.yahoomail.com
Learning Objectives
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.
System of Linear Equations
A system
of linear equations is two or more linear equations that are being solved
simultaneously.
Solution of a System
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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 said consistent, give yourself a pat on the back!
If you said independent, you are correct!
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 said inconsistent, you are right!
If you said independent, you are correct!
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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 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!
<!--[if !supportLineBreakNewLine]-->
<!--[if !vml]--><!--[endif]--> 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:
2x-3y=3 Plug in 3 for x and 1for y
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:
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?????
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:
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?????
Three ways to Solve Linear Equations in two variables
three ways to solve systems of linear
equations in two variables:
Solve by Graphing
Step 1: Graph the first equation.
(3, 1) and (0, -1)
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.
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.
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 !vml]--> Example 2:
Solve the system of equation by graphing.
Step 1: Graph the first equation.
x-intercept
The x-intercept is (3, 0).
y-intercept
The y-intercept is (0, 3).
Find another solution by letting x = 1.
<!--[if !vml]--><!--[endif]-->
x-intercept
y-intercept
The y-intercept is (0, -1).
Find another solution by letting x = 2.
Another solution is (2, 1).
Solutions:
Step 3: Find the solution.
We need to ask ourselves, is there any place that
the two lines intersect, and if so, where?
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).
Solve by Substitution method
Step 1: Simplify if needed.
This would involve things like removing ( )
and removing 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.
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.
Step 4: Solve for
the remaining variable .
Solve the equation set up in step 3 for the
variable that is left.
If your variable drops out and you have a FALSE
statement, that means your answer is no solution.
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.
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.
<!--[if !vml]--><!--[endif]--> Example 1: Solve
the system of equations by the substitution method:
<!--[if !vml]--><!--[endif]-->
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.
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)
Step 5: Solve for second variable.
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.
Example 5:
Solve the system of equations by the substitution method:
<!--[if !vml]--><!--[endif]-->
Step 1: Simplify if needed.
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.
Note how the second equation is
already solved for y. We can use that one for this step.
<!--[if !vml]--><!--[endif]-->
Y=-3x+4 *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)
Wait a minute, where
did our variable go????
Step 5: Solve for second variable.
Since we did not get a value
for x, there is nothing to plug in here.
There are no ordered pairs to
check.
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.
Step 2: Multiply one or both
equations by a number that will create opposite coefficients for either x ory 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.
Step 3: Add equations.
Add the two equations together.
Step 4: Solve for remaining
variable.
Solve the equation found in step 3
for the variable that is left.
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.
Example 7:
Solve the system of equation by the elimination method:
<!--[if !vml]--><!--[endif]-->
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.
Again, you want to make this as
simple as possible.
Step 3: Add equations.
Step 4: Solve for remaining variable.
-x=-10
Step 5: Solve for second variable.
You can choose any equation used
in this problem to plug in the found x value.
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.
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The following are webpages that can
assist you in the topics that were covered on this page.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut19_systwo.htm
This webpage will help you with solving linear equations in two variables.
http://www.purplemath.com/modules/systlin1.htm
This webpage goes over the basic definition of systems.
http://www.purplemath.com/modules/systlin2.htm
This webpage will help you with solving linear equations in two variables by
graphing.
http://www.purplemath.com/modules/systlin4.htm
This webpage will help you with solving linear equations in two variables by
substitution.
http://www.purplemath.com/modules/systlin5.htm
This webpage will help you with solving linear equations in two variables by
addition/elimination.
<!--[endif]-->
you should be able to:
|
A system
of linear equations is two or more linear equations that are being solved
simultaneously.
Solution of a System
<!--[if !supportLineBreakNewLine]-->
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 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 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!
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:
|
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 said dependent, you are correct!
The graph below illustrates a system of two
equations and two unknowns that has an infinite number of solutions:
|
<!--[if !supportLineBreakNewLine]-->
| (3, 1) and (0, -1) |
Let’s check the ordered pair (3, 1) in the first
equation:
|
2x-3y=3 Plug in 3 for x and 1for y
2(3)-3(1)=3 True Statement
3=3
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:
|
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:
|
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:
 |
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 Linear Equations in two variables
three ways to solve systems of linear
equations in two variables:
|
Solve by Graphing
Step 1: Graph the first equation.
|
(3, 1) and (0, -1)
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.
|
Step 1: Graph the first equation.
|
x-intercept
|
The x-intercept is (3, 0).
y-intercept
|
The y-intercept is (0, 3).
Find another solution by letting x = 1.
|
x-intercept
|
y-intercept
|
The y-intercept is (0, -1).
Find another solution by letting x = 2.
|
Another solution is (2, 1).
Solutions:
 |
Step 3: Find the solution.
|
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).
|
Solve by 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.
|
Step 4: Solve for
the remaining variable .
|
Solve the equation set up in step 3 for the
variable that is left.
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.
|
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 ory 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.
|
Step 3: Add equations.
|
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:
|
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:
|
Step 3: Add equations.
|
Step 4: Solve for remaining variable.
|
-x=-10
x=10
Step 5: Solve for second variable.
|
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.
|
|
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.
|
<!--[if !supportLineBreakNewLine]-->
The following are webpages that can
assist you in the topics that were covered on this page.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut19_systwo.htm
This webpage will help you with solving linear equations in two variables. http://www.purplemath.com/modules/systlin1.htm This webpage goes over the basic definition of systems. http://www.purplemath.com/modules/systlin2.htm This webpage will help you with solving linear equations in two variables by graphing. http://www.purplemath.com/modules/systlin4.htm This webpage will help you with solving linear equations in two variables by substitution. http://www.purplemath.com/modules/systlin5.htm This webpage will help you with solving linear equations in two variables by addition/elimination. |
