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factor 25-y^2

I would just square root both right? :P

difference of two squares!

(x+5)(x-5) ?

Yep!

Thanks :)

Now,
Simplify \(\Large \frac{3x^3}{6x^2}\)

\[\frac{ 3*x*x*x }{ 3*2*x*x }\]
\[\frac{ x }{ 2 }\]

-_-

You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0

is that ok?

Most teachers will take half the points away if you forgot that, or some of these conditions.

O_O

That was over copied!
Simplify \(\Large \frac{x(x^2+6)}{x^2}\)

(x^2+6) / x

Condition?

\[x \neq1\]

You cancelled x, so the condition is x\(\ne\)0

It's what you cancelled should not equal to zero.

x\(\ne\)1 is for cancelling (x-1)

are we good?

Yes

ok, ready for the next one!

mhm

Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)

Remember to specify the conditions if you cancelled any factor.

\[\frac{ (pâˆ’1)(pâˆ’1) }{ p^2-1}\]

Continue! There are factors for the denominator.

diff of two squares!

In the worst case, use the quadratic formula!

\[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]

Excellent, you can now finish off. Tag on the conditions whenever you cancel!

\[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]

** -1

x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0

is that clear how we got x\(\ne\)1 ?

yes

ready for another?

no more of these ;-;

2

you get the idea, but a little too fast ! :(

Remember m-4 = -(4-m), and 3/6=1/2

... and the conditions!

x /= 1/3 , 3/6

So the final answer is
-1/2, where m\(\ne\)4

yessss

34.
simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)

\[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]

anything else to do or to write?

(note: the factoring is correct! Well-done!)

nope thats all : P

ok :)

Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)

i need help ;-;

1. any common factors (in the denominator)?

no

2. Is it diff. of 2 squares?

no

3. Is it a perfect square?

no

So now can you complete it?

Shall we go back to perfect squares?

\[\frac{ 1 }{ 3x-5 }\]

Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3

>.<

So the answer is:

1/3x-5
x/= 5/3

Yep, that's correct!

We check that with an example
Factor 81x^2- xy +64y^2.

Last term = 64 y^2, square-root = ?

3 :/ ?

Last term = 64 y^2, square-root = ?

8y

Good.
What's twice the product of the square-roots?

huh ?

We need to multiply together the square-roots (of the first and third terms), and double it.

512? :/

im so confused :/

|dw:1433718072207:dw|

|dw:1433718524311:dw|

|dw:1433718696561:dw|

Thats so much more better :)

I understand this a bit more :o

Good! An image is worth 1000 words, I was told!

xD

So we'll be back on track!?

yes

i cant do it without a calc! :O

It's not the answer we want, just the process. I'll do that for you.

\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}

hmm ok

\[\frac{ (x-3) }{ 3(xâˆ’1)(2x+3)(xâˆ’3) }\]

correct so far?

exactly, continue!

\[\frac{ 1 }{ 3(x-1)(2x+3) }\]
x/= 3

Almost.
The last step is all correct.
There was a mistake in the first step:

Oh, actually, you were right, I was wrong.
The final answer is good! Congrats!

O_o

Do you still remember how to do a division of fractions?

I'd say: flip the fraction after the \(\div\) sign.

Do you agree?

Can we keep doing multiplication ?

Sure! Another one coming up

\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)

Simplify!

\[\frac{ (c+8)(câˆ’8) }{ 4cc(c+1)(c+8) }\]

correct so faR?

Excellent, go on!

Remember you cancelled a "c" already. Keep track of that.

Can you finish it off ;-;

Is the last part ok for you now?

yes :)

Another multiplication?

nope ;-;

yes

We would do the same with rational functions.

Gimme a second to type it up.

simplify
\(\Large \frac{x}{x+6} \div \frac{x+3}{x^2-36}\)

\[\frac{ x }{ x+6 }\times \frac{ (x+6)(x-6) }{( x+3) }\]
\[\frac{ x(x-6) }{ (x+3) }~~~~~x \neq6\]

Perfect!!!! Can't be better!

REALLY? :O

I did that without a calc :O

Oops, what about the previous ones ! LOL

smh

Ready for another?

yes!: )

Simplify
\(\Large \frac{45x^3-9x^2}{x} \div \frac{6(x-5)}{2}\)

\[\frac{ 3x(5xâˆ’1) }{ x-5 }\]

Wow, what did you eat for supper?

xD

Another?

No i have this : ) next case!

k. coming right up!

11.5 Adding and Subtracting with like denominators
=================================

Like denominators mean that the denominators are either the same, or just a simple multiple.

So it's a matter of adding the numerators, like:|dw:1433721592963:dw|

ok

now gimme a minute.

Add
\(\Large \frac{x+2}{x} + \frac{3x-2}{x}\)

\[\frac{ 4x }{ x }\]

...and the next step is...

the answer would be 4

Because the x's cancel right?

Yes, ...and...

and??

x\(\ne\)0
"conditions whenever cancelling a factor in the denominator".

i knew it! >:(

haha!
Another?

Next case!

\(\Large \frac{3x-4}{x-4} - \frac{2x}{x-4}\)

Just curious!

i knew it was coming :P
\[\frac{ 1x-4 }{ x-4}\]

recall 1y = y

1

and..

and??

x\(\ne\)4

The conditions are the only difficult part in these examples!

ugh

|dw:1433722130469:dw|

multiplication right?

Nope, perimeter would be the sum of all four sides.

Next case!

Please!

We only have two more cases.

I dont like this one ;-;

im not even sure where to start .-.

|dw:1433722487080:dw|

can you finish it?

\[\huge~\frac{ 2(xâˆ’2)(x^3+4x^2âˆ’12x+4) }{ (x+2)(xâˆ’2)(xâˆ’2) }\]

wow, not like that.
The addition had already been done, because the denominators are the same.

All you need to do is to cancel (x-2) and give the answer!

|dw:1433722913558:dw|

>.<

Got it

good!

another?

no ;;

Once you can find the common denominator readily, rest is relatively straight-forward.

Makes sense :)

So that's good, we'll proceed with an example.

Simplify
\(\Large \frac{x+1}{5} + \frac{2x}{6}\)
Can you first try to find the common denominator?

|dw:1433726621112:dw|

Do you agree?

Yess

\[\huge~\frac{8x+3}{15} \]

Exactly, so we go with the same idea for the other rational expressions.

Now do you still remember equivalent fractions?

yes

for example,
\(\Large \frac{9}{5x} = \frac{? }{40x^4 }\)

8?

It's basically cross multiplication,
\(\Large ? =40x^4 ~\frac{9}{5x} = 72x^3\)
Dose that make sense?

so
\(\Large \frac{9}{5x}= \frac{72x^3}{40x^4}\)

is that ok? We need to use this step later on.

yes its fine

Ok, try to find the sum in its simplest form:
\(\Large \frac{1}{x^2} + \frac{2}{x}\)

\[\huge~\frac{2x+1}{x^2}\]

Yes, that correct!

\[\large~\frac{ 3 }{ x }\]

cannot do that, because x and 1 are not like terms, so there is nothing to factor out.

oh right

Add and simplify if possible \(\Large \frac{1}{x+1} + \frac{1}{x-1}\)

\[\huge~\frac{2x}{\left(x+1\right)\left(x-1\right)} \]

Wow, are we wasting time?

Try this: add
\(\Large \frac{x-5}{x+5} + \frac{x+2}{x-2}\)

Sorry, subtract!

\(\Large \frac{x-5}{x+5} - \frac{x+2}{x-2}\)

Use FOIL.

\[\huge~\frac{14x}{\left(x+5\right)\left(x-2\right)}\]

\(\Large \frac{x}{x-10} + \frac{x+4}{x+6}\)

wait i cant do this one .-.

When there is no common factor, you only have to "kind of" cross multiply.

\[\frac{2x^2-40}{\left(x+6\right)\left(x-10\right)} \]
i have it uptil this part

|dw:1433728624180:dw|

Yes, it's correct, but need to factor out the 2 to give 2(x^2-20) at the numerator.

Another one?

nuu next case!

ok. Have you done rational functions?

p.678

11.7 Rational Functions
==============

Havent done that

Sure?

yes

ok, good.
Together with the calculator(s), we have covered everything!

._. lol thanks for all the help :)

You're welcome! :)