College Algebra
Tutorial 49: Solving Systems of
Linear Equations in Two Variables
Learning Objectives
Introduction
Tutorial
System of Linear Equations
In this tutorial, we will be looking at systems that have only two linear 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.
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.
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:
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:
*True statement
Now, let’s check (3, 1) in the second equation:
*True statement
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:
*True statement
Finally, let’s put (0, -1) into the second equation:
*False statement
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 by Graphing
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.
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.
If it makes at least one of them false, you need to go back and redo the problem.
y-intercept
Find another solution by letting x = 1.
Solutions:
Plotting the ordered pair solutions and drawing the line:
y-intercept
*Inverse of mult. by -1 is div. by -1
*y-intercept
Find another solution by letting x = 2.
*Inverse of mult. by -1 is div by -1
Solutions:
Plotting the ordered pair solutions and drawing the line:
The answer is yes, they intersect at (2, 1).
The solution to this system is (2, 1).
y-intercept
*y-intercept
Find another solution by letting x = 1.
Solutions:
Plotting the ordered pair solutions and drawing the line:
*Inverse of mult. by -1 is div. by -1
*x-intercept
y-intercept
Find another solution by letting x = 1.
Solutions:
Plotting the ordered pair solutions and drawing the line:
The answer is no, they do not intersect. We have two parallel lines.
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.
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!!).
This will give you one equation with one unknown.
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.
If it makes at least one of them false, you need to go back and redo the problem.
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:
*Inverse of add 3 is sub. 3
(4, 3) is a solution to our system.
Multiplying each equation by it's respective LCD we get:
*Mult. by LCD of 2
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:
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.
The answer is no solution.
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:
*Variable dropped out AND true
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.
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 MethodTo 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.
The variable that has the opposite coefficients will drop out in this step and you will be left with one equation with one unknown.
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.
If it makes at least one of them false, you need to go back and redo the problem.
Multiplying each equation by it's respective LCD we get:
*Mult. by LCD of 40
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:
*y's
have opposite
coefficients
I choose to plug in 10 for x into the first simplified equation (found in step 1) to find y’s value.
*Inverse of add 30 is sub. 30
*Inverse of mult. by 2 is div.
by 2
(10, 24) is a solution to our system.
Rewriting the second equation we get:
*Everything is lined up
So we do not have to multiply either equation by a number.
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.
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}.
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:
*x's
and y's
have opposite coefficients
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.
The answer is no solution.
Practice Problems
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?
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.
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on March 25, 2011 by Kim Seward.
All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.