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This video covers the process of FOILing two binomials. Multiply: First, Outer, Inner, Last. A trick is given to help you remember the order.

Welcome to Your Tutor Online video lessons. Today we are going to look at how to FOIL. FOILing is what you do when you multiply two binominals together. FOIL is an acronym which stands for First, Outer, Inner, and Last.

When we talk about the first terms, we are going to look at each respective parenthesis. The first term in each parenthesis is the first term. The outer terms are for the outside of the entire thing, our 2x and our negative 5 are both our outside terms. The inside is just the opposite, so the two closest to each other. And then last, the two terms that are last in each of the parenthesis. So, 3 and negative 5.

And now we just follow our acronym in order: First, Outer, Inner, Last. First, we will multipy together our two first terms, our 2x and our x which gives us 2x squared. Next is outer, 2x and negative 5. 2x times negative 5 is negative 10x. Our inside terms are 3 and x which gives us positive 3x. And our last terms 3 and negative 5 gives us negative 15.

Now, almost always you will end up with your two middle terms being like terms so you just want to combine those two as your final answer. 2x squared, combine our two like terms (-10x and positve 3x) gives us negative 7x and minus 15.

Here’s a tip for you to remember to do the FOILing in the correct way. If you look back up here, how I drew the lines, I have almost a little smile face. Here is the smile, the nose, and two eye-ball looking things. If you get the smile face at the end that means you drew your FOIL correctly and you multiplied the terms together correctly.

Alright, I am going to show you how to draw this face one more time so you have a good check for when you have to do FOIL on your own. We are going to follow in order: First, Outer, Inside, Last. For the first term, we will start above the problem. Connect the first two terms with an arc. So in this example, 3x and 2x and that will give us our first eye brow. Now the outside, we will connect our two outside terms 3x and 1, which gives us our smile. The inside terms, the 2 and the 2x, gives us our nose. And our last terms, negative 2 and positive 1, give us our last eyebrow. And now if you did it correctly you will see a face in here. And now you know you did your problem correctly.

I hope you found this lesson useful. And, as always I’m glad to accept your donations. I am asking $1 per lesson. That is the cheapest tutoring that is out there. You can head over to my website at YourTutorOnline.com and click the donate button at the top of the page. I want to thank everybody for all of their comments they leave over at YouTube, keep them coming. I appreciate the comments and questions. And if you need more help, I am available for private tutoring. Just head on over to my webpage at YourTutorOnline.com and click on the tutoring link at the top of the page for more information. You could all help me out by writing a review of my podcast at PodcastAlley.com or write a review in iTunes. I will see you all next time, class dismissed.

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This video covers the basics of matrices and how to add them together. I am trying my hand at Adobe Flash. Let me know how you like this new style by clicking “Comments >>” below.

Welcome to Your Tutor Online video lessons. Today we will look at matrices and matrix addition.

A matrix is a way to organize numbers. It has rows and columns. This gives us our dimensions. In this example, this matrix is a 2 by 3 matrix because it has two rows and three columns.

In order to add matrices they must have the same dimensions. Here, we have two matrices that have 2 rows and 2 columns each. All we have to do is add up the corresponding numbers in each matrix and put that answer in a new matrix in the same spot. So, we’ll start with the numbers in the first row, first column; add those together and put that answer in the first row, first column. We’ll repeat with the number in the first row, second column and so on for each number until you’ve got your answer.

Also in this series:

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

Welcome to Your Tutor Online video lessons. This is the last of a 4 part series on systems of equations. In this video we will look at how to solve systems of equations by substitution.

Substitution just means you’ll find out what one variable is equal to and you plug it into the other equation. In this first example we already see that x is the same thing as 2y minus 3. So that means where-ever we see an x in the other equation we can put in this for it instead. And that way we will only have ys to deal with and be able to solve the equation.

So I’ll begin to re-write the second equation: 3y plus 2 x, and instead of x we are going to put in what we know x is equal to. X is the same thing as 2y minus 3; is equal to 8.

Now that we only have one variable, we can solve for y. 3y and distribute the 2; plus 4y minus 6 is equal to 8. Combine like terms and get 7y minus 6 is equal to 8. Add 6 to both sides. 7y is equal to 14. And divide both sides by 7. We find out that y is equal to 2. We are going to take this answer we got for y and plug it back into either one of the original equations to solve for x.

I chose to plug this back into the first equation: x = 2y - 3. So we are going to take our y is equal to 2, put it where we see the y and go ahead and solve for x. 2 times 2 is 4. Subtract 3 and x is equal to 1. And now we know both solutions. X is equal to 1 and y is equal to 2.

Sometimes a variable will not already by isolated, or by itself, so you have to work with an equation to get one of the variables by itself. In this example the second equation could easily be turned into one where y is by itself if we just subtract 3x from both sides. So it becomes y = -3x + 5. And we can take that, just like we did before, and plug that into the other equation where-ever we see a y.

4x - 2; plug in the y; -3x +5 is equal to 10. Now all we have are x’s and we can solve for x. Distribute the -2; So we get plus 6x minus 10 is equal to 10. And we’ll continue this on another page. We’ll combine our like terms the 4x and the 6x leaves us with 10x and at the same time I’ll add 10 to both sides. 10 plus 10 is 20. We’ll divide both sides by 10 to get x by itself and have our answer for x is 2. Now that we know what x is we can go back to one of the original equations, plug in 2 where we see an x and solve for y.

Here I plugged in 2 into the first equation and we will solve for y. 4 times 2 is 8 minus 2y is equal to 10. We’ll subtract 8 from both sides and be left with a -2y on the left, a +2 on the right. Divide both sides by a -2 and y is equal to -1.

That’s all there is for solving systems of equations by substitution. They get a little more complex especially when the variables are not already by themselves. But follow this process and you’ll be able to solve any of them.

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
• 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.