BAM!!!
Mr. Tarrou.
Well, in the next couple of videos we are going to take our skills of factoring, review
our skills of finding common denominators and adding or subtracting fractions.. or rational
expressions, and start solving some equations.
In this video we are going to be dealing with just fractional coefficients, actually fairly
straight forward, and then we are going to work in a couple of word problems that have
the same design in there algebraic equations... but they are word problems.
Then we will do another video where we have some algebraic expressions in both the numerator
and the denominator of the fractions.
That is really where you have to start needing to use all of those algebra skills we have
been learning in the last few videos.
About adding, subtracting, and multiplying algebraic fractions.
What we are going to do with our rational equations, or inequalities.. one of these
examples is going to be an inequality, is simply try and get rid of the fractions as
fast as possible.
That means multiplying by the LCM or Lowest Common Multiple to get rid of the fractions
as quickly as possible to simplify the problems.
So we have a denominator here of 15, 15, and 6.
I can deal with those all the way through the problem, or I can multiply both sides
of the equation by something that will cancel out those denominators, the LCM.
What number does 15 and 6 both go into?
You can do prime factorizations and break it down like that.
But hopefully at this point, knowing our multiplication tables really well, say 6, 12, 18, 24, 30.
Oh, 15 goes into 30.
So we are going to multiply both sides of this equation by 30.
And on this first problem I am going to show this before we actually do any cancellation,
I will show all the multiplication just so you can see it.
But you can probably do this in your head as well.
We have positive 30 times 4x^2 over 15... equals positive 30 times -11 over 15 times
x... and a positive 30 which is what we are multiplying by.
I am going to take care of this negative right now, times negative 1/6.
Ok.
Well, 15...
Excuse me 30 divided by 15 is equal to 2.
30 divided by 15 is again equal to 2.
And 30 divided by 6 is equal to 5.
So now that we have cancelled out all of those denominators we can handle the rest of this
problem, hopefully with using the skills you just learned in some past sections.
We have 2 times 4 which is 8 x squared.
We have 2 times negative 11 which is negative 22x.
And we have 5 times negative 1 which is -5.
We are going to get everything over to one side of the equation because it is a quadratic.
We have a term with a degree of 2, and we have a term with a degree of 1.
So we are not just going to get x by itself.
That is not going to be possible.
With these quadratics, where you have this setup where you cannot just get n, x, or ax
alone, you have to set the equation equal to zero.
And solve it by factoring or the quadratic equation, or something.
Let's see what we can do.
We are going to move the -22 with addition.
Every time we go over... from one side of the equal sign or inequality sign to the other
side, we must do the inverse math operation.
So we are going to add both sides by 22x... add both sides by 5... and we get 8x^2 plus
22x plus 5 is equal to zero.
Now that leading coefficient is quiet a bit larger than one, and that means that we are
not going to necessarily to easily factor this quadratic trinomial.
So we can solve this using the Quadratic Formula if you have learned that already.
Hopefully you have.
x is equal to negative b, or the opposite of b, plus or minus the square root of b squared
minus 4ac all over 2a.
I however have made a problem that I know factors.
So we are going to take that leading coefficient and that constant and we are going to multiply
them together.
8 times 5 is equal to 40.
We are going to be looking for factors of that product that add up to the middle term.
Well, 2 times 20 is equal to 40... and 2 plus 20 is, when we add them, equal to 22.
That is our middle coefficient.
So we are going to rewrite the middle term.
Rewrite this whole expression as a polynomial with four terms instead of a trinomial.
Our two middle terms are written with these coefficients.
Then we are going to factor that by grouping.
These first two terms share a common factor of 2x.
8x^2 divided by 2x is 4x.
And 2x divided by 2x is 1.
So just plus 1.
These two terms both share a common factor of 5.
We are going to divide that out.
You will notice that we have a common factor within our two new terms.
That is 4x+1.
We are going to divide out or factor out that 4x+1.
And write 4x+1 times 2x+5 is equal to zero.
And then...
Now that we have this in factored form, we are going to solve this equation by factoring.
We just did a bunch of videos about factoring.
Did we not?
So I wonder why that was:)
We have two factors being multiplied together to give us a function that is equal to zero.
The only way to multiply and get an answer of zero is if one of those factors are equal
to zero.
So we have 4x+1 equals zero and 2x+5 equals zero.
We need to solve those out individually and I am starting to run out of room here.
Did not realize how much room this example was going to need.
Let's just make some more room.
We are going to take 4x+1 and have it equal zero.
We are going to subtract both sides by 1.
Divide both sides by 4 and get negative...
not negative one-half... how about negative one-fourth is one of our solutions.
And take the 2x+5 and set it equal to zero.
Subtract both sides by 5.
Divide both sides by 2.
That answer is negative five halves.
Now it is very very important with rational equations that you take these answers and
you plug them back into the original problem and make sure that they work.
For one that is always a good practice to make sure that your answers are indeed correct.
And with rational equations you have a good chance of something being an extraneous solution....
Dividing by zero.
You don't want one of your solutions to allow your original equation to have a division
of zero in it somewhere.
You want to make sure you plug these back into your original problem and make sure that
both sides are actually equal.
That is really not going to be an issue with this problem because we don't have a variable
in a denominator.
So it is not possible for any of our solutions to make either one of those denominators equal
to zero.
And let me just make sure that I am not making any mistakes.
Good!
We have the same answers that I have in my notes.
Let's get to the other problem here.
I am going to need to erase these so make sure you get those answers down.
Ok.
For this example over here we only have one fraction on the left hand side and one fraction
on the right.
So we could solve this with cross multiplication.
And the step I am going to show which is multiplying both sides by the lowest common multiple is
going to achieve the same thing in about the same amount of time.
I just want you to know that one fraction equal to one fraction, cross multiplication
is a valid method of solving that equation.
We, just to be consistent, are going to multiply the entire equation..
I am going to show this a little bit different than the other one.
I am going to just show one huge parenthesis around the entire equation.
And we are going to multiply by 12.
Why 12?
Because 12 is divided by 12 is going to cancel out that denominator.
And 12 is divisible by 4.
So it is going to cancel out that denominator out as well.
A fraction bar is one big grouping symbol.
Right?
It says take all of this on top and divide it by what is in the denominator.
So we have one term on the left like I just said and one term on the right.
We are going to do this distributive property and only have to do it twice because there
are only two terms.
So all of this is divided by 12 and I am multiplying by 12, so that is going to cancel out completely.
We get x times x+13.
Over on the right hand side, 12 divided by 4 is 3.
So we have 3 times x+8.
We are going to distribute through these parenthesis and get x squared plus 13x equals 3x plus
24.
We have a squared term and a single degree term so it is not going to get this equation...
um... where you just have x equals a number.
And that is it.
So we are going to need to get this equation equal to zero and then either factor or use
the quadratic formula.
This one will be factorable as well.
We are going to bring this 3x over with subtraction.
I guess I probably should be showing that.
So let's do this.
We are going to do minus 3x from both sides.
We get x^2 plus 10x equals 24.
Bring the 24 over the same way.
It is a positive 24 so we are going to subtract 24 from both sides.
Now we have a quadratic with the leading coefficients of one.
So this factoring will be quite easy.
Factors of 24 that subtract to be 10?
Positive 12 times negative 2 is negative 24, and positive 12 minus 2 is equal to 10.
So those are the factors that we want.
We get factors of the first term in the first two blanks.
Factors of the last term, the ones that we want are positive 12 and negative 2 in the
last two blanks.
We have two factors being multiplied together to give us zero.
So each on of those... or one of those needs to be zero.
So x+12 is equal to zero and x-2 is equal to zero.
Both of these can't be true at the same time, but if either one of them are true then we
are multiplying by zero and thus we will get zero and it is a solution.
So subtract both sides by 12.
We get x equals -12.
That is one solution.
Add 2 to both sides and get x=2.
That is the other solution.
When we plug those values in, either the -12 everywhere there is an x or the positive 2
everywhere there is an x, we will be getting two sides of the equation that are equal.
We have one more example.
BAM!
Here is our last example.
x over 4 minus x-3 over 2 is greater than or equal to 2x+1 over 5 minus 1.
Except for the fact this has four terms instead of two or three like the other examples, the
only real difference... and it is kind of a big one, is we have an inequality in the
middle somewhere and not an equal sign.
The steps for solving this is exactly the same as solving an equation.
The only difference, as by now I hope you know, is when you multiply both sides of the
inequality by a negative number, or you divide both sides of the inequality by a negative
number, you have to change the direction of the inequality.
So let's get going.
We are going to look at our denominators.
We have 4, 2, and 5.
I like to take the biggest denominator...
well take one of the numbers... and start running through the multiples of that number
until you recognize one where all of those values will go into it.
So let's take 5.
Ok.
5, 10.. 2 divides into 10 but not 4.
15, 20 Ok.
So 20 divided by 4 is 5.
20 divided by 2 is 10.
And 20 divided by 5 is equal to 4.
So if I multiply everything by 20, the Lowest Common Multiple, I will cancel out all of
those fractional coefficients and make my like a little bit easier.
So we are going to do that.
We are going to multiply these two terms by 20.
This is a positive 20 so I don't need to worry about changing the direction of the inequality.
Instead of showing quite so many steps, I am going to try to do this in my head a little
bit.
Hopefully you can keep up.
20 divided by 4, because this is for you.. not me... 20 divided by 4, that is equal to
5.
So 5x...
Now I am going to keep this minus here... 20 divided by 2 is equal to 10.
So negative 10, and because it is negative I am going to keep these parenthesis in here.
So negative 10, that is the 20 divided by 2.
There is the minus sign... times x-3... is greater than or equal to... 20 divided by
5 is 4.
I think I am going to go ahead and keep these parenthesis for the binomial in here.
4(2x+1)... and then minus not one.... but 20 times -1 is negative 20.
Now let's distribute those parenthesis.
Make sure you do take care of that sign as you do this multiplication.
So we have 5x... minus 10x... plus 30...
All greater than or equal to 8x plus 4 minus 20.
Ok.
We have...
Oh.
For a change we don't have a second degree equation here so we are going to be able to
just get x by itself.
We are going to add like terms.
We have 5 minus 10 which is giving negative 5x.
And we have 4 minus 20 which is negative 16.
Now I have a variable, and x on both the left and right hand side.
If I add both sides by 5x, that is fine.
And I am not going to have to worry about changing the direction of the inequality.
But I do see students make more mistakes when they move their variables to the right hand
side.
Not that is an incorrect step.
But it is the type of mistakes...
I see more mistakes when move their variables to the right as opposed to when they move
them to the left.
So even I am going to need to change the direction of this inequality here in another step coming
up, I am going to go ahead and subtract both sides by 8x and move my variable to the left.
So negative 5 and negative 8 is negative 13... plus 30 is greater than or equal to negative
16.
I am going to subtract both sides by 30.
And that will give us -13x is greater than or equal to... that will be -46.
Ok... so now I have got -13x is greater than or equal to -46.
I need to solve for x.
I am going to move, or undo that multiplication of negative 13 by dividing.
I am going to divide by -13.
This is going to cancel out.
We do have x.
We are dividing by that negative number, so we are going to need to change the direction
of that inequality.
Now it is less than or equal to.... and not much is going to happen here if you are keeping
your answers in exact format.
The negatives are just simply going to cancel out.
We have 46 divided by 13 and that is our answer.
And that means I am done.
So I am Mr. Tarrou.
BAM!
Go Do Your Homework:)