Thursday, December 18, 2008
Wednesday, December 17, 2008
Solving Rational Equations
Hi! Today we had two visitors who monitored us in class. Mr. Kuropatwa went over yesterday’s work and also gave us new questions to work on. During the workshop, we has to reach upto 15 slides but we only got to about 9 slides. I will TRY my best to explain everything so you can understand it:)
First, we should all know how to multiply binomials by this time.
For example:
And when its factored it looks like this:
This should all be familiar to us.
When factoring trinomials, you get the ones at the top and when you reverse it you get multiplying binomials. It is like undoing the expressions.
Now Mr. K introduced us an easier and faster way to multiply two digit numbers.
For example:
41 x 31 = ?
First you multiply the two 1st digits
4x3 which equals to 12 then multiply the second digits which are1x1=1
so, so far you have 12 and 1. .leave a blank in between 12__ 1
now to find out what the blank is, you need to do a rainbown multiplication
the rainbow multiplication would be 1x3=3 and 4x1=4 and 3+4 = 7
the middle number would be 7.
The Fast Multiplication relates to the first part because of how the procedures are done.
Heres a picture that might help you understand
Now the second part that we went over is balancing equations.
Ever since i have learned balancing equations form the past, I have
been transposing. .I guess it was the faster way of doing it and easier and I still get the right answer but today in class Mr. K. taught us something new. .
In this type of questions or any questions we DO NOT transpose because we don't just throw numbers from on side to another, we balance it.
to find the x. .
we make zero pairs so since there is x + 3 on the left side and 2 + 3x on the right side.
Add a negative x on the left side and add negative two on the right side.
Right now it isn't balanced so what we do on one side we do on the other
So we need to add -x and -2 on each side to balance the equation--there we go BALANCE EQUATIONS
The next question was. .
First of all find the greatest common factor or GCF. . not LCD because we are dividing to get rid of the denominator.
In this case the gcf is 4 because 4 can be devided by 2 and 4 evenly. When you have found the gcf, you have to multiply it by every term to balance it.
On the first term 4÷2=2 so 2 stays there since it cannot be reduced anymore. Now on the last term, you can reduce 4÷4=1 so it is reduced. REM
EMBER! "REDUCE" not cancell
Now since everything has been reduced you can multiply the rest
Finally. .
The gcf on this equation is x.
However . Make sure to state the restrictions after finding the gfc.
Now multiply x by every term to balance it. On the first term the x's are going to reduce so you're left with a four.
Just multiply every term by x and you will be left with 4= -4x.
To get the x by itself, divide both sides by -4.
JUST REMEMBER DO NOT FORGET TO STATE RESTRICTIONS OR ELSE YOU WILL LOSE A MARK !!!
Mr. K asked us if this was a rational equation, since it can be in fractions it is shown to be rational.
Please feel free to comment on this. Let me know if I have made any errors or if you want to share something and I'll add to it. If you still don't understand, talk to someone who knows best. Thanks and I hope you understand my scribe. .HAPPY HOLIDAYS!!!*
THE NEXT SCRIBE WILL BE ÿâssiÑⁿ♣ !!!!!!!!!!!
Today's Slides: December 17
Tuesday, December 16, 2008
Simplifying Rational Expressions
The class was kind of slow today, we just did some questions on simplifying rational expressions.
Some notes:
- We don't do subtraction in High school, we add a negative number.
- Give everybody a home
- Expressions are different from Equations:
the denominator is cleared out.
This is one of the expressions Mr. Kuropatwa asked us to do:
First, we factor it out.
Before we lose a mark, we should state the restrictions now.
Then we find the common denominators which is:
Now we should multiply each expressions with the number one, in another form, and it should give us the common denominator "because we built it that way".
,
Then we just add the numerators because we already have the common denominators.
This would be the final result.
But... in some expressions we could still further by factoring the numerator and reducing any common factors, but in this case we can't. Mr. K taught us a way to find out if we can still factor an expression, here's how:
We multiply the leading coefficient by the constant.
Then we look for the factor of that number that will add up to 9 which is the middle term.
In this case we can't and that makes it a PRIME number!
That's all I could say so see yah!
And Levi would be the next scribe...
Today's Slides: December 16
Monday, December 15, 2008
Operations On Rational Expressions
So it would be look like....
So the restriction would be look like...
c.)
And then after you solve this.. you will get this answer:
It would look like this...
Today's Slides: December 15
Sunday, December 14, 2008
Rational
Step 1: Factor the numerator and the denominator.
Step 2: Express each common factor pair in the numerator and denominator as 1.
Step 3: Simplify and state any restrictions on the variable.
Here are some examples:
Well....The next scribe will be Jigsaw....Jan