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This video explains how to factor a quadratic equation with a leading coefficient. (an equation that looks like ax^2 + bx + c).

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.

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.

This lessons looks at how to simplify radical expressions which are not perfect squares or cubes. I include an example of square root, cube root, and 4th root. I show a technique to simplify n-root expressions. Also I look at how to simplify variables in radical expressions.

Welcome to Your Tutor Video lessons. Today we’ll look at how to simplify radicals.

A radical is anything with a radical symbol which looks like this. It can be square root, cube root, fourth root, or any numbered root. If there is no number in the symbol then its a square root, otherwise there will be a tiny number here to show you what root it is. In this example it is cube root. These expressions need to be simplified like anything else in math.

Lets look at this example: radical 54 or square root of 54. The easiest way I have found to simplify radicals is to do a factor tree for the terms inside the radical. Now this only works if everything inside the radical is multiplied together or you only have one term.

First draw two lines out from the number or term inside the radical. And on this line we are going to write any two numbers that multiply together to give us our originial. I’m going to choose 9 and 6. We are going to repeat the process until we have nothing left except for prime numbers. 3 times 3 is equal to 9 and 3 times 2 is equal to 6. Now we are left with only prime numbers here at the bottom, 2s and 3s.

Now we want to look and see which root we are dealing with. Here it is the square root or 2nd root because there is no number in the symbol itself. Since its the second root, we are looking for groups of two of the same numbers. Here I see two-3s. When you get a group, you can write the group number outside of the radical, in this case its 3. Then we’ll go ahead and write our radical and write everything thats left over inside, multiplied together: 3 times 2. Go ahead and multiply those together: 3 radical 6, and thats our simplified expression.

You can look for shortcuts as you do your factor tree. As you get more practice with this type of problem, you will know that 9 is a perfect square and so it’s square root can automatically come out of the radical. For cube roots you are looking for perfect cubes and so on according to whatever radical you happen to be dealing with. Now its going to get harder as you get into higher radicals and you probably won’t have high powers memorized, but thats okay you can just do your factor tree when in doubt.

Lets look at how this works for a cube root. We’ll use the example cube root 135. We are still going to go the factor tree just like we did in the previous example. 5 times 27 is equal to 135. 5 is a prime number so we will just leave that alone. 27 is 3 times 9 and 9 is 3 times 3. Notice we could of had a shortcut there if you knew that 27 was a perfect cube.

This time we are going to look for groups of three because we are dealing with the cube root. Here you can see three-3s. So that means a 3 is allowed to come out of the radical. So I write 3, the radical symbol with its cube root and whatever is left over that didn’t find a group of three, which is 5. So this radical simplified is 3 cube roots of 5.

Lets look at how to do radicals with variables in them. For a coefficient we are still going to do a factor tree. 48 is 12 times 4. 12 is 3 times 4. 3 is a prime number. 4 is 2 times 2. And this 4 is 2 times 2. This time we are going to be looking for groups of four because we are dealing with the fourth root. So here, I found four-2s. So the two is going to come out and a 3 will stay inside.

Now lets look at our variables. This part is a little tricky. We are going to look at each of our exponents for our variables. We are going to take our root number and divide it into each of the exponents. The number of times that the root goes into the exponent, will be our new exponent on the outside of the radical and the remainder will be the new exponent that stays behind inside the radical.

Now we can get started on our answer. We know a 2 comes out so I’m going to write that. I’ll leave some space, draw my radical symbol with is the fourth root. I know a 3 is stuck inside. Now I can start dealing with the variables.

4 goes into 6 one time so x to the first power is on the outside. 4 goes into 6 once and there are 2 left over, so its x squared left on the inside. 4 does not go into 2 any so a y does not come out of the radical. 4 goes into 2 zero times with 2 left over so we have a y squared left on the inside. 4 goes into 8 two times so we have z squared on the outside. 4 goes into 8 twice with no remaineder so there is no z left on the inside. And this is our final answer: 2 x z squared times the fourth root of 3 x squared y squared.

Thanks for watching Your Tutor video lessons, now I know today was a little bit of a tricky topic so if you have any questions head over to the blog and leave a question at www.YourTutorOnline.com And also, if you haven’t been there yet be sure to check out the quizzes I’m going to put up after each lesson and see how much you understand. Thanks for watching, class dismissed.

This lesson shows how to solve equations involving absolute value.

Welcome to Your Tutor Online video lessons. Today we will learn how to solve equations that have an absolute value in them.

As a review, absolute value refers to how far away a number is from zero. To put it simply, the absolute value makes a number positive no matter what. Three is three units away from zero. Negative four is four units away from zero. So the absolute value of three is three, and the absolute value of negative four is four. Three stays positive, number four becomes positive.

What if there is a variable inside the absolute value sign? Well, in our order of operations, we want to treat the absolute value sign just like we would parentheses. So simplify everything inside them first, and move everything you can to the other side. In this example, 3 times the absolute value of x plus 2 is equal to 9, we can divide both sides by 3 and we are left with the absolute value of x plus 2 is equal to 3. When you get the absolute value all by itself on one side, you are done with the simplification.

Now we need to split this equation into two separate equations. For the first one, simply drop the absolute value signs. We have x plus 2 is equal to 3. For the other equation drop the absolute value signs and this time make the other side of the equation negative. Or if its already negative make it positive, just flip the sign of the other side of the equation.

Solve each equation that you came up with. On the left x is equal to 1 and on the right x is equal to negative 5. So for absolute value equations we will have two solutions, In this case x is equal to 1 or x is negative 5.

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 lessons suggestions send an email to podcast[at]yourtutoronline.com Thanks for watching, class dismissed.

This lesson covers how to identify, name and label points and lines.

Welcome to Your Tutor Online video lessons. Today we will learn how to identify, name and label points and lines.

A point does not take up any space, but so that we can see it we draw a dot. Points are labeled with capital letters. You call a point by its letter, and is read “point A.”

Two points make up a line, and a line extends forever in both directions. We can represent that by drawing arrows on either side. To name a line pick any two points that are on the line. Here we have points A and B. Lines are also labeled with capital letters, so this is called “line AB.” It can be any letters on the line at all. If we had another point here, C, it can be “line AC”, “line CA”, “line CB”, all those are fine labels for this line. The symbol for a line is the two capital letters with a line over top, just remember to draw the arrows. We read this “line AB.” You can also label a line with a lower case letter. For example we can say this is “line n.”

Points can be collinear or non-collinear. Collinear just means that the points are all on the same line. A,B, and C are all collinear points. Points D, E, and F are non-collinear because you cannot draw a line that will connect all the points.

A line segment is part of line, has end points, and does not go on forever. Line segments can be measured. You name a line segment by its end points. For example here we have line segment AB. The symbol for a segment is the capital letters of the two end points with a bar on top. The order here doesn’t matter. Segment AB is the same as segment BA.

Because line segments can be measured they can be compared with one another. If two lines segments are the same lenght they are called congruent. The symbol for congruent is an equal sign with a funny hat on top. An up and down dash can also represent congruency. Segment AB is congruent to segment BC. When this happens we know that B is the midpoint of segment AC because it divides that segment into two equal parts.

When we are comparing things, such as line segments, we use the word congruent. When we are comparing measurements we use the word equal. The measurement of segment AB is equal to the measurement of segment BC.

A ray has an endpoint and extends forever in one direction. To name a ray always start with the end point and then pick one other point on the ray. This example can be called ray AH. The symbol for a ray is going to be the capital letters of the end point (first) and then any other point with a tiny ray above it. The ray above it will always point to the right regardless of the way the ray actually points.

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 lessons suggestions send an email to podcast[at]yourtutoronline.com Thanks for watching, class dismissed.