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Yep.
Do you know direct and inverse variation?

no ._.

oh ok :)

ok with direct?

yes :)

ok for inverse variation?

Yes i get that :)

Can we move on the rational expressions?

* to

yes

A rational expression is a polynomial divided by another polynomial.

Example,
(5x+3)/(2x+1)
is a rational expression.

i just foil right?

Before you do foil, did you notice anything?

we can cancel

Yes, only as long as (x-1) does not equal zero!

Yes :)

So we say:
\(\large \frac{(4x+3)(x-1)}{(x-1)}=4x+3\) if x\(\ne\)1

ok with that?

Yes :D

We cannot tell off-hand if there is any common factor, so what do we do?

We use the wheel thingy

Good, or "factorize" is the other word for the wheel thingy!
Can you do that, please?

|dw:1432867194095:dw|
\[(4x^2-4x)+(3x-3)\]
\[4x(x-1)+3(x-1)\]
\[(4x+3)(x-1)\]

\[\frac{ (4x+3)(x-1) }{ (x-1) }\]
(x-1) cancel out.
Final answer:
\[ \huge 4x+3\]

So the answer is again 4x+3 with x\(\ne\)1.

I dont like that :/

the condition?

Yep

Yes, it complicates life a little, but that's the only difference from fractions.

Are we good so far (except you don't like the condition! :) )

yes :)

ok so far?

yes.

ok so far?

yes :)

Can we just skip those :P i know those

What about with unlike denominators, are you ok with those?

no...

yes

yes

Now for rational exprssions:
\(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\)

What would be the common denominator? (do not FOIL, otherwise we need to factorize afterwards.)

x+4

ok?

To this we have to add the conditions that x-2\(\ne\)0 and x+4\(\ne\)0

ok so far in making equivalent fractions?

im so confused and lost T_T

NO!! T_T

Yesterday we went through 11.2 to 11.6.
Now it's time to do some practice!

next class

:( your way out!

Just kidding, sure, later!

O_O

Just remebered i have a thing to do.... oops cant work :P

Ok, I'll grab you later, ok?

BWAHAHAHA XDDD :3 yr so innocent xS

im free :# we can math :P

ok! I was O_o

R U good with 11.3?

:/

Want a practice to make sure?

yes

BTW, were the correct answers in the quiz all between 11.2 and 11.3?

no :/

So you did get some in other sections, that's good!

ummm no

i thought i did great on it but i ended up failing

;-; can we start with something easier :(

Ok, simplify
\(\large f(x)=\frac{x^2+x-6}{(x+3)(x-4)}\)
:)

|dw:1432933988455:dw|
IDK T_T

|dw:1432934285205:dw|
Which pair works?

3 -2 .-.

(x+3)(x-2)

Exactly. Are you able to complete the answer now?

\[\frac{ (x-2)}{ (x+4) }\]

no

The condition that (x+3)\(\ne\)0.
Teachers are waiting to jump on that! :(

So the complete answer is
(x-2)/(x-4) with x\(\ne\) -3

ok so far?

I hate that

ye the rest is fine :)

Want another one similar?

no :(

T_T !

ok fine another one

|dw:1432935523037:dw|
\[\frac{(x-2)(x+3) }{ (x-5)(5+3)}\]
Final answer
\[\frac{ (x-2) }{ (x+5) }\]

oh...oh

O_O ?

Remember the last problem? transcription and something missing???

T_T i dont know this!!! >:(

:)

We were on 11.3

All you have to remember is if you get candies, you pay!

candies ~ cancel

:( i hate that

Sorry, we have to stick to rules!
Now
can you simplify
(x-2)(x+4)/(x+4)
?

(x-2)

You cancelled (x+4) right?

yea....

So we write
(x-2).....

thats what i did .-.

You were close:
(x-2) for x\(\ne\)2
is the complete answer.

* for x\(\ne\) -4

>:(

ok, let's try another:
simplify (x-4)(x+2)/(x-4)

(x+2)

Ahhh >:( T_T

Now try this:
simplify (2x+5/2)(x+3)/(2x+5/2)

No no more of these T_T

x+3
;-;

Can we please skip these T_T

Simplify
(2x+3)(x-1)/(x-1)

2x+3 x/= 1??????

Are you ok now?

Yes

Ready for 11.4?

Yes

I iz here

11.4 is on rational expressions and not radical expression, I suppose. Can you confirm?

yea lol rational is right

ok, try this:
simplify \(\large \frac{x-2}{x+3}\times\frac{x+3}{x-4}\)

(x-2)(x+3)/(x+3)(x-4)
\[\huge \frac{ x-2 }{ x-4 }\neq 3\]

correction:
\(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq -3\)
because we don't want x+3=0

Now, try
Simplify
\(\large \frac{x-2}{5x+6}\times\frac{2x-3}{x-2}\)

(x-2)(x-2)/(5x+6)(2x-3)
T_T
Lets do a diffrent one! :|

ok try this:
Simplify \(\large \frac{x^2-5x+6}{5x+6}\times\frac{5x+3}{x-2}\)

(x-3)(x-2)(5x+3)/(5x+6)(x-2)
\[\frac{ (x-3)(5x+3) }{ (5x+6)(x-2) }\neq-2\]

>_<

Simplify \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x-6}\)

O_o

ok i kinda get it

ye

It seems better now.
Are you good with multiplication and division of rationals?

kind of :/

yeah lol my hoe work ill put up a problem from that : )

Sure! It will be less boring.

xD
\[\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ 3x^2-9x }{ x^2+7x+10 }\]

So what would be your first step?

factor?

* always be...

ooo right flip

\(\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ x^2+7x+10 }{ 3x^2-9x }\)

Oops too small, but I can still read it.

lol ok so now we factor right?

\(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)

Actually on second look, would the original question be a multiplication instead of a division?

You can start with writing them separately in two separate numerators.
After that, we cancel.

But it seems that there is not much to cancel unless the original question was a multiplication.

its division :/

Sorry, I take it back. There is at least one that we can cancel.

huh :?

How would i write out the factors?

For the ones in the numerator, you can put them where they are now.

If you want, we can factor the left numerator together.

\(\large x^2+2x-15\)
=\(\large (x- \ )(x+ \ )\)

Good job! Speed of a bullet!

like:
(x-3) means x\(\ne\)3

\[\frac{ (x+5)^2 }{ 3x }\]

Yes, but there are two "3x" , so you write 9x^2 at the bottom.

>:(

\[x \neq-3~~~~~~~x \neq2\]

Does it have to be written like that? :/

It would be the simplest, simpler than what I would have done.

\[\frac{ (x+5)^2 }{ 3x },x \neq2,x \neq-3\]

That's ok too, but do remember it's 9x^2 at the bottom, because we had two 3x left at the bottom!

oo right >_< oops

\(\large\frac{ (x+5)^2 }{ 9x^2 },x \neq2,x \neq-3\)

Excellent, you got one done, and it wasn't a simple one like my (boring) ones.

XD

Got other ones to try?

\[\huge \frac{ x^2-64 }{ 3x^3 }by~(8-8)\]

what's "by" O_o

Sure it's (8-8) ? (equals zero!) O_o!

\[\huge \frac{ (x+8)(x-8) }{ 3x^3}\div \frac{ 1 }{ (x-8) }\]

Ok! I think you did the division twice (you flipped, and you kept the division sign!)

Is it
\(\huge \frac{ (x+8)(x-8) }{ 3x^3}\times \frac{ 1 }{ (x-8) }\)

And you factored the top right-away, ;) ?

yes :) xD

Excellent!

What's your answer then?

Bet you this one comes out in the quiz.

\[\huge\frac{ 1(x+8) }{ 3x^3 },x \neq -8\]

Another one done, with 100%.

:) MORE!!!:D

yes please! More, more....

You pick :P ;)

I like your examples better, they are less boring.

ok, I'll pick one.

LOL xD

Simplify \(\large \frac{x^2+6x+9}{x^2-9}.\frac{3x-9}{x^2+2x-3}\)

\[\frac{ (x+3)(x+3) }{ (x+3)(x-3) }\times \frac{ 3x-9 }{x+3)(x-1) }\]

Wow!

3x-9 = 3(x-3)

Did you use the perfect square and diff. of two squares?

or you factored all three?

Excellent.
Now you cancel, and note the conditions as you cancel.

(x-3) and (x+3)^2 will be cancelled top and bottom, leaving
\(\large \frac{3}{x-1}\), x\(\ne\)-3,1

* \(\large \frac{3}{x-1}\), x\(\ne\)-3,3
You don't have to worry about x+3 twice.

ready for another?

yes! :)

Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8} \div \frac{2x^2-x-3}{x^2-3x-4}\)

\[\frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (2x-3)(x+1) }{ (x+4)(x-1) }\]

Great, just you need to flip, and I think bottom left should read (x-4)(x+1) before flipping.

:(

* bottom-right

wait so what should it look like :. ?

The answer should be very simple, almost the simplest possible, plus three conditions.

*four conditions

all of them cancel out! T_T

That's good, so what's the answer? (it's not zero).

\[\huge x~\neq-4~~~x~\neq2~~~~~x~\neq1~~~~~????\]

Almost,
compare 5/5=1
x-4\(\ne\)0 means x\(\ne\)4, etc.
so \(\large 1, x\ne -2,-1,3/2,4\)

Continue with these "easy" ones, or some addition?

T_T

i wanna move onto something more challenging :)

Like addition and subtration?

*subtraction

ye

ok,
Simplify \(\large \frac{4}{x^2-16}+\frac{3}{x^2+8x+16}\)

Step 1: factorize, then we can find the lowest common multiple (LCM)

or the common denominator.

\[\huge \frac{ 4 }{ (x+4)(x-4) }+\frac{ 3 }{(x+4)(x+4) }\]

:/

Excellent!

*one

x+4

x+4 does not contain x-4, and x+4 twice.
Hint: the common denominator here has three factors.

;-; idk

Similarly for the second term:
\(\large \frac{3\color{blue}{(x-4)}}{\color{blue}{(x-4)}(x+4)(x+4)}\)

= \(\large \frac{4x+16+3x-12}{(x-4)(x+4)(x+4)}\)
= \(\large \frac{7x+4}{(x-4)(x+4)(x+4)}\)

and that's the answer.

Did you follow all the steps?

If you get stuck, tell me where.

oh ok i get it :)

ok =)

Try another one?

No T_T

Do you have anything from your school notes?

Example or exercises?

no o.o

ok, shall we take a break/

mmmmmmmmmmmmmmmm k :)

No lets continue

Do you have some examples for the easy ones (like denominators)?

No. I'd prefer you give the problems... :/

Ok, the boring ones!

Simplify \(\large \frac{4x+3}{x^2-16}+\frac{3x+4}{x^2-16}\)

\[\frac{7\left(x+1\right)}{\left(x+4\right)\left(x-4\right)} \]

Excellent!

Simplify \(\large \frac{x^2-3x+2}{2x-4}-\frac{x-1}{2x-4}\)

\[\frac{\left(x-3\right)\left(x-1\right)}{2\left(x-2\right)} \]

Oh yes, I thought the difference is not factorable! lol Great job!

v_v next

k

\(\large \frac{2x+3}{x^2+4x+4} +\frac{x^2+2x+1}{(x+2)^2}\)

;-; not this one! Lets skip it! :D

\[\huge\frac{ 2x+3}{ (x+2)(x+2)}+\frac{ (x+1)(x+1) }{ (x+2)(x+2) }\]

for me?

\[\huge\frac{ 2x+3(x+1)^2}{ (x+2)(x+2) }\]

Watch out:
\(\huge\frac{ 2x+3\color{red}{+}(x+1)^2}{ (x+2)(x+2) }\)

Keep going, you're almost there!

You need to expand the (x+1)^2 using FOIL or identities and then add.

\[\frac{ 2x+3+x^2+1x+1x+1 }{ (x+2)^2}=\frac{ x^2+4x+4 }{ (x+2) }=\frac{ (x+2)(x+2 }{ (x+2)(x+2)}\]

Great!
So after cancelling you have a numeric answer with a single condition:

\[x \neq2\]

The answer is 1 (5*5/(5*5)=1)
and the condition is x+2\(\ne\)0, or x\(\ne\)-2

Remember we don't want what we cancelled to be zero.

got it.

Try the unlike denominators now or later?

I'll start with this:
Add \(\large \frac{2}{9}+\frac{7}{12}\)

Please show work!

No lets skip this section i dont like this section

I IZ HERE

hold on! have to find the question!

ugh no i put everything away ;-; lets just use yours :(

mathmate i thin your pm's are lagging

*think

omg

omg

First step: flip.

\(\large \frac{2x^2+x-6}{x^2-2x-8}\times \frac{x^2-3x-4}{2x^2-x-3}\)

Next, factor, one piece at a time, and show intermediate steps, please!

\[\huge \frac{ 2x^2+x-6 }{ x^2-2x-8 }\times \frac{ x^2-3x-4 }{ 2x^2-x-3 }\]

\[huge(x+4)(x-3)\]
\[\huge(2x^2+4x)(-3x-6)\]
\[\huge2x(x+2)~~~~~-3(x+2)\]

Then proceed with grouping:
2x^2+4x -3x-6 = 2x(x+2) -3(x+2) = (2x-3)(x+2)

Good job, continue please!

[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]

\[(2x^2+2x)(-3x-3)\]
\[2x(x+1)~~~~~~~-3(x+1)\]
\[(2x-3)(x+1)\]

You mean
\[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]

\[[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ (2x-3)(x+1) }\] \]

so far so good!

Now the finishing touch!

the answer is 1 .-.

right, but not finished yet!

omg no T_T the conditions

Didn't your teacher ask you to specify the conditions, or is it just I?

no he doesnt make us list those :/

Never?

nope :/

Even in the quiz? Is it multiple choice?

;-;

So the four conditions are:
2x-3\(\ne\)0
x+2\(\ne\)0
x-4\(\ne\)0
x+1\(\ne\)0

\[x \neq 3/2~~~~~x \neq-2~~~x \neq4~~x \neq-1\]

So you can write them together as:
x\(\ne\)-2,-1,3/2,4

Excellent, thank you! :)

>:(

Ready for another one, simpler, and no conditions!

yes!

\(\large \frac{4+2x}{x^2-4}.\frac{x^2-4x+4}{x-2}\)

Hint: x^2-4 is the same as x^2-2^2, difference of 2 squares.

\[\frac{ 4+2x }{ (x+4)(x-4) }\times \frac{ (x-2)(x-2) }{(x-2) }\]

and x^2-4x+4 is a perfect square, if you recall!

You can do better with the first expression (on the left, both top and bottom.

Remember bottom was x^2-2^2

2(2+x)

The bottom is (x+2)(x-2)

answer is 1

I think it's 2. Can you check?

yh its 2 ;o

Good.
Want to do rational addition/subtraction (more difficult) or rational equations?

None of the above. xD

ok, it's getting late!

LOL ok we can continue tommzzz

*tommrow

ok!