Your Tutor Online

Also in this series:

• Lesson 1: Introduction
• Lesson 2: Graphing
• Lesson 3: Elimination (current page)
• Lesson 4: Substitution

Welcome to Your Tutor Online Video Lessons. Today we continue our series on solving systems of linear equations. Today we will look at how to solve those equations by elimination.

For the elimination method you will simply add the two equations together in a way that will cancel out one of the variables. In our example here we will just go ahead and add the two equations together as we see them. The ys will cancel out. 3 plus 2 gives us 5 x. Negative 1 plus positive 1 cancels out; and 10 plus 20 is 30. Now that the ys are canceled out we can solve for x. In this example x is equal to 6. Then we want to take our solution, x = 6, go back to one of the original equations and plug it in. So we plugged 6 in for x and we will solve for y. 2 times 6 is 12, plus y, is equal to 20. Take away 12 from both sides and we are left with y = 8. Now we have both answers to our system of equations. X = 6 and y = 8.

Sometimes you need to change one of the equations before adding them together in order to get a variable to cancel out. In this example if we multiply the second equation by negative 2, the xs will cancel out. When you multiply an equation by a number, you need to multiply every single term by that number. So now we get -2x + 4y = 22. And we can just copy over the top equation exactly as we see it. 2x + 3y = 13. Now when we add these two equations together the xs will cancel out. Positive 2 plus negative 2 is zero. 4 and 3 is 7y, and 13 plus 22 is 35. We can divide this new equation by 7 on both sides and we will solve: Y = 5. Again we are going to take this same answer, plug it into one of our originals and we will get the answer for x. X -2 times 5 = -11. X - 10; -2 times 5 is -10 = -11. Add 10 to both sides and x is -1.

For our last example we will see that sometimes you have to multiple both equations by a factor in order to get something to cancel out. Here, none of the coefficients are multiples of the other. We need to multiply both equations in order to match up any of the coefficients. I am going to try to eliminate the x. In order to do that I will multiply the 2x by 5 and the 5x by -2 to get a 10 and a -10 to cancel out. 2x times 5 becomes 10x; 3y times 5 becomes 15y; and 16 times 5 is 80. For this equation we are multiplying by -2 so we get -10x; +8y; all that is equal to +12. Add both of these together. The 10s cancel out: +10-10 is zero. 15 + 8 = 23y. 80 + 12 = 92. We will divide both sides by 23 to get the y all by itself. And y = 4. Same process as before, we are going to take our answer for y and plug it into an original. 2x + 3y; y here is 4, is equal to 16. 2x + 12 = 16. Subtract 12 from both sides. 2x = 4; and finally 4 divided by 2. X = 2.

I hope you found this lesson useful. If you have any questions leave a comment on the blog at www.YourTutorOnline.com If you have any lesson suggestions send an email to podcast@yourtutoronline.com Thanks for watching, class dismissed.

Also in this series:

• Lesson 1: Introduction
• Lesson 2: Graphing (current page)
• Lesson 3: Elimination
• Lesson 4: Substitution

Welcome to Your Tutor Online video lessons. Today we continue with the second video in our four part series on solving systems of linear equations. In this lesson we’ll look at solving systems by graphing. Now, this is the easiest, but also the most time consuming method of solving systems. You have to re-arrange the equations into a graph-able form; set up a graph; plot some points; and then find out if the intersect or not. All you need to do is graph the two equations you are given on the same coordinate plan. If you need a refresher on this you can look over my previous lesson on graphing equations.

As discussion in the introduction: if the lines intersect then the point of intersection is the solution, if they are parallel then there is no solution, and if they happen to be the same line then there are infinite solutions.

Lets look at just one example: 3x + 2y = 6 and 4x - 3y = 12. I prefer to work with slope-intercept for so I’ll go ahead and re-write each equation.

So our new equations, after we put them in slope-intercept form, are y = - (3/2)x + 3 and y = (4/3)x - 4. Now we just need to graph these two equations on the same coordinate plane.

We’ll start with y = - (3/2)x + 3. We’ll start at the y intercept, +3, so go up 3 on the y axis. And then we’ll use the slope to get a couple other points on the line. Down 3 and over 2. And we will go one more time. That’s enough points, now we just need to connect the dots and we have our line.

For the other equation we are going to go to the y intercept first, -4, and put a dot. Are we are going to rise 4 and run 3. And we’ll connect those dots.

The two lines intersect, and the point where they intersect will be our solution. In this case it’s not whole numbers but that’s okay, you can give an estimate. And this looks to be 2.5 and -0.5. Your teacher may have his or her own requirements for rounding or guessing if this particular situation comes up.

I hope you found this lesson useful. If you have any questions leave a comment on the blog at www.YourTutorOnline.com If you have any lesson suggestions send an email to podcast@yourtutoronline.com Thanks for watching, class dismissed.

This lesson is the first of a four part series about systems of linear equations. Here, I define what a system of equations is, and the three types of solutions. Future lessons will cover graphing, elimination, and substitution.

Also in this series:

• Lesson 1: Introduction (current page)
• Lesson 2: Graphing
• Lesson 3: Elimination
• Lesson 4: Substitution

Welcome to Your Tutor Online video lessons. This is the first of a four part series on solving systems of linear equations. This first lesson will serve as an introduction.

A system means a group, or more than one. And the word linear here, refers to a line, and also means we will be dealing with two equations which each have two variables in them. Normally they will be x and y. We need to work with both equations at the same time to get a solution. Here’s an example of what a system of equations problem looks like. 3x minus 4y is equal to 2 and 7x plus 3y is equal to 4. Now, using the ways I’ll show you in future lessons you will be able to work with these two equations and get a solution for x and y.

There are three types of solutions to this kind of problem. You can have one solution, no solutions, or infinite solutions.

You have one solution when the two lines intersect. The point where the lines intersect is your solution. You can give the answer as an ordered pair, in this case “(4,2)” or you can split up the coordinates and say that x is equal to its coordinate which is 4 and y is equal to its coordinate which is 2.

A system has no solutions if the lines are parallel or never intersect. This can be seen clearly with graphing. Or, prior to graphing if the two lines have the same slope but not the same y intercept then they are parallel and have no solution. You can simply write “no solution” or this symbol, a zero with a slash through it, which means the same thing.

There are infinite solutions if the two lines you are dealing with are actually the same line. Here is an example. I have a blue line and a green line right on top of each other. This can happen if the first equation you are given is a multiple of the second or vice versa. We’ll take a closer look at that in upcoming lessons. These two are the same line, so they’ll intersect at every single point. So, we say there’s an infinite number of solutions.

There are three ways to solve systems of linear equations: By graphing, elimination, and substitution. Each of these will be the focus of the next three lessons. Be sure to visit www.YourTutorOnline.com to see them when they become available. And, as always, if you have any questions leave a comment underneath any of the lessons. Thanks for watching, class dismissed.

I will be putting up 4 new video lessons. They will all look at how to solve systems of linear equations. Click on a link below to jump to that lesson.

• Lesson 1 will serve as an introduction
• Lesson 2 will look at solving by graphing
• Lesson 3 will cover elimination
• Lesson 4 will wrap it up with substitution

Lesson 1 will be posted tomorrow morning, bright and early.