@mathmate

- pooja195

@mathmate

- Stacey Warren - Expert brainly.com

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- mathmate

Yep.
Do you know direct and inverse variation?

- pooja195

no ._.

- mathmate

Direct variation is a function that varies directly with x.
For example
X Y
1 2
2 4
3 6
4 8
5 10
is a direct variation. Notice that Y/X is always the same ratio (of 2).
so far so good?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- pooja195

oh ok :)

- mathmate

On a graph, direct variation is a straight line passing through the origin.
|dw:1432866053517:dw|
Here f1(x), f2(x), f3(x) are all direct variations, but g(x) is not, because it does not pass through the origin.

- mathmate

ok with direct?

- pooja195

yes :)

- mathmate

For inverse variations, they work differently.
Y is the inverse (or reciprocal) of x. This way, X*Y is a constant (same number).
Example:
X Y
1 2
2 1
3 2/3
4 1/2
5 2/5
6 1/3
...
notice the product of X*Y is always 2.
the graph looks like this
|dw:1432866347703:dw|

- mathmate

ok for inverse variation?

- pooja195

Yes i get that :)

- mathmate

Can we move on the rational expressions?

- mathmate

* to

- pooja195

yes

- mathmate

A rational expression is a polynomial divided by another polynomial.

- mathmate

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

- mathmate

The sum of two rational expressions is also a rational expression, just like the sum of two fractions is still a fraction.

- mathmate

Even with one single rational expression, we can simplify , for example, what would be
\(\large \frac{(4x+3)(x-1)}{(x-1)}\) ?

- pooja195

i just foil right?

- mathmate

Before you do foil, did you notice anything?

- pooja195

we can cancel

- mathmate

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

- pooja195

Yes :)

- mathmate

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

- mathmate

ok with that?

- pooja195

Yes :D

- mathmate

So far, we've been lucky. Both the numerator and denominator contained (x-1) as a common factor. What if we are given the following to simplify:
\(\large \frac{4x^2-x-3}{x-1}\) ?

- mathmate

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

- pooja195

We use the wheel thingy

- mathmate

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

- pooja195

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

- mathmate

Excellent!
Now you replace the numerator of the rational expression with what you've got, what do you get?
\(\large \frac{4x^2-x-3}{x-1}= ?\)

- pooja195

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

- mathmate

Whenever you cancel out in a rational expression, you need to ADD a condition that what you cancel out is not zero, so you need to add x\(\ne\)1 (or x-1\(\ne\)0).

- mathmate

We don't do that in fractions because we almost always work with non-zero factors.
If we had a fraction 5*0/4*0, cancelling the common factor 0 would give 5/4 as the answer, which is not the same as 0/0.

- mathmate

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

- pooja195

I dont like that :/

- mathmate

the condition?

- pooja195

Yep

- mathmate

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

- mathmate

So whenever you need to simplify a rational expression, factorize as much as you can, and cancel as much as you can. Add one condition whenever you cancel a factor (unless the same factor appears twice)
Example:
(x+3)(x+1)^2/((x+1)^2(x+5))
=(x+3)/(x+5) if x\(\ne\)-1

- mathmate

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

- pooja195

yes :)

- mathmate

If we multiply rational functions, it's the same as in fractions. Again, if we cancel, we need to specify conditions!
\(\large \frac{x+5}{x-2} * \frac{x-2}{x+3} = \frac{x+5}{x+3} \) if x\(\ne\)2

- mathmate

ok so far?

- pooja195

yes.

- mathmate

For divisions, it's the same as in multiplication, except that we need to flip the divisor and multiply.
Example:
\(\large \frac{x+5}{x-2} \div \frac{x+5}{x+3} = \frac{x+5}{x-2} * \frac{x+3}{x+5} =\frac{x+3}{x-2}\) if x\(\ne\)-5

- mathmate

ok so far?

- pooja195

yes :)

- mathmate

Now for adding and subtracting with like denominators, it would be just like in fractions. Should I give an example anyway?

- pooja195

Can we just skip those :P i know those

- mathmate

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

- pooja195

no...

- mathmate

ok.
But back to those with like denominators: after adding or subtracting, you will still need to factorize and look for factors to cancel AND add a note/condition. Is that ok?

- pooja195

yes

- mathmate

Ok, now +/- expressions with unlike denominators.
Just like 1/3+2/5
common denominator is 3*5=15
so we'll make equivalent fractions before adding
= 5/15 + 6/15
now add
=11/15
ok?

- pooja195

yes

- mathmate

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

- mathmate

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

- pooja195

x+4

- mathmate

We need the product of the two denominators,
(x-2)(x+4), so that we can make equivalent expressions.
\(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\)
=\(\large \frac{(x+1)(x+4)}{(x-2)(x+4)}-\frac{(x-2)(x-2)}{(x+4)(x-2)}\)

- mathmate

ok?

- mathmate

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

- mathmate

ok so far in making equivalent fractions?

- mathmate

The rest of it is adding two rational expressions with like denominators.
After addition/subtraction, you need to factorize and cancel expression (and add conditions if applicable).

- pooja195

im so confused and lost T_T

- mathmate

@pooja195 xD

- pooja195

NO!! T_T

- mathmate

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

- pooja195

next class

- mathmate

:( your way out!

- mathmate

Just kidding, sure, later!

- mathmate

@pooja195
Time to work!

- pooja195

O_O

- pooja195

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

- mathmate

Ok, I'll grab you later, ok?

- pooja195

BWAHAHAHA XDDD :3 yr so innocent xS

- pooja195

im free :# we can math :P

- mathmate

ok! I was O_o

- mathmate

R U good with 11.3?

- pooja195

:/

- mathmate

Want a practice to make sure?

- pooja195

yes

- mathmate

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

- pooja195

no :/

- mathmate

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

- pooja195

ummm no

- mathmate

Now try to simplify:
\(\large f(x)=\frac{x^2+x-6}{x^2-x-12}\)
Do not forget to mention the condition, if any

- pooja195

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

- pooja195

;-; can we start with something easier :(

- mathmate

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

- pooja195

|dw:1432933988455:dw|
IDK T_T

- mathmate

Remember, when the product (-6) is negative, the two numbers you're looking for have different signs. Does that help?

- mathmate

|dw:1432934285205:dw|
Which pair works?

- pooja195

3 -2 .-.

- pooja195

(x+3)(x-2)

- mathmate

Exactly. Are you able to complete the answer now?

- pooja195

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

- mathmate

Almost, just a transcription error (should be x-4 at the bottom)
AND
missing very important information...
Do you know what's missing?

- pooja195

no

- mathmate

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

- mathmate

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

- mathmate

ok so far?

- pooja195

I hate that

- pooja195

ye the rest is fine :)

- mathmate

Just have to remember that whenever you cancel a term, it comes with a condition.
Like when you buy candies, you always pay! :)

- mathmate

Want another one similar?

- pooja195

no :(

- mathmate

T_T !

- pooja195

ok fine another one

- mathmate

ok, here's another one. Notice that the numerator is the same as the previous, so you are allowed to reuse your own work this time.
Simplify \(\large \frac{x^2+x-6}{x^2-2x-15}\)

- pooja195

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

- mathmate

oh...oh

- pooja195

O_O ?

- mathmate

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

- pooja195

T_T i dont know this!!! >:(

- mathmate

Good up to this:
\(\frac{(x-2)(x+3) }{ (x-5)(x+3)}\)
After that, the final answer after cancelling is
\(\frac{(x-2)}{ (x-5)}\) with the condition that x\(\ne\)5.

- mathmate

@pooja195

- pooja195

:)

- mathmate

We were on 11.3

- mathmate

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

- mathmate

candies ~ cancel

- pooja195

:( i hate that

- mathmate

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

- pooja195

(x-2)

- mathmate

You cancelled (x+4) right?

- pooja195

yea....

- mathmate

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

- pooja195

thats what i did .-.

- mathmate

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

- mathmate

* for x\(\ne\) -4

- pooja195

>:(

- mathmate

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

- pooja195

(x+2)

- mathmate

Almost, it would be (x+2) for x\(\ne\)4
because if x=4, the expression/function is 0/0, and that does not have a value.
So our answer of (x+2) is valid as long as x\(\ne\) 4

- pooja195

Ahhh >:( T_T

- mathmate

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

- pooja195

No no more of these T_T

- mathmate

I am looking forward to see you do one completely by yourself, because I won't be there to help you when you do your quiz!

- pooja195

x+3
;-;

- mathmate

The answer is (x+3) for 2x+5/2\(\ne\)0
Remember when 2x+5/2=0, the expression becomes 0/0 and is undefined. So to avoid this situation, we specify that 2x+5/2\(\ne\)0

- pooja195

Can we please skip these T_T

- mathmate

Sorry, math is a cumulative knowledge. If we don't do it right, all the following topics will be wrong!
I am asking you to show me at least once that you will follow a cancellation with a condition.

- mathmate

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

- pooja195

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

- mathmate

yes, put if, for, for all, or whenever... to indicate that it is a condition.
so 2x+3 for all x\(\ne\) 1

- mathmate

Are you ok now?

- mathmate

I will skip this step and move on to 11.4. But remember that in all the other topics, whenever you cancel, you still have to write the condition, else the answer is wrong without it.
Are we good?

- pooja195

Yes

- mathmate

Ready for 11.4?

- pooja195

Yes

- mathmate

@pooja195

- pooja195

I iz here

- mathmate

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

- pooja195

yea lol rational is right

- mathmate

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

- pooja195

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

- mathmate

That's the idea!
We write the condition separate though,
\(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq 3\)

- mathmate

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

- mathmate

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

- pooja195

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

- mathmate

You need to cancel the (x-2) at the top with the (x-2) at the bottom, the will give (2x-3)/(5x+6) for all x\(\ne\) 2

- mathmate

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

- pooja195

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

- mathmate

you cancelled the x-2 at the bottom, so x-2\(\ne\)0, or x\(\ne\)2
The answer will then read:
\(\frac{ (x-3)(5x+3) }{ (5x+6)} \ for\ all\ x\neq2\)

- pooja195

>_<

- mathmate

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

- mathmate

You'll need to invert the second term to change division to multiplication:
=\(\large \frac{x-3}{5x+6}\times \frac{5x-6}{x-2}\)

- mathmate

oops the question is wrong. It is meant to have 5x+6 top and bottom!
\(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x+6}\)
=\(\large \frac{x-3}{5x+6}\times \frac{5x+6}{x-2}\)

- pooja195

O_o

- mathmate

Now we cancel the common factor 5x+6 (as long as 5x+6\(\ne\)0.
so we write
=\(\large \frac{x-3}{x-2}\) for all 5x+6\(\ne\)0
=\(\large \frac{x-3}{x-2}\) for all x\(\ne\)-6/5

- pooja195

ok i kinda get it

- mathmate

@pooja195 we were at 11.4 are you ready?

- pooja195

ye

- mathmate

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

- pooja195

kind of :/

- mathmate

ok, we'll go through all sections, and then come back to those which you feel wobbly.
BTW, do you have some examples from school, we can work on those.

- pooja195

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

- mathmate

Sure! It will be less boring.

- pooja195

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

- mathmate

So what would be your first step?

- pooja195

factor?

- mathmate

That's ok, but I would prefer to put the second term upside down, so we will always working with multiplication.

- mathmate

* always be...

- pooja195

ooo right flip

- mathmate

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

- mathmate

Oops too small, but I can still read it.

- pooja195

lol ok so now we factor right?

- mathmate

right, lemme fix the div sign, but you can go ahead and factor.
Remember the first step is to take out the common factors of each expression, ex. in 3x^2+6x. Take out 3x.

- mathmate

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

- pooja195

Ok this is the part where i get kinda confused....you know how theres like two seperate equations on top? They both can be factored but how would i write it out?

- mathmate

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

- mathmate

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

- mathmate

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

- pooja195

its division :/

- mathmate

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

- mathmate

So you can go ahead and factor, as though they are two separate terms.
We can combine them after cancelling.

- pooja195

huh :?

- mathmate

I mean you can proceed with this:
\(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)

- pooja195

How would i write out the factors?

- mathmate

The bottom one on the right should take out 3x as well, so you put
\(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x(x-3) }\)

- mathmate

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

- mathmate

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

- mathmate

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

- pooja195

ok
|dw:1433012589434:dw|
|dw:1433012632806:dw|
\[\frac{ (x+5)(x-3) }{ 3x(x+2) } \times \frac{ (x+5)(x+2) }{ 3x(x-3) }\]

- mathmate

Good job! Speed of a bullet!

- mathmate

can you now post me the factors that you can cancel, and write down the condition associated with each one?

- mathmate

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

- mathmate

You can write all the factors on top as one single numerator, and similarly for the denominators.
Looks like this:

- pooja195

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

- mathmate

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

- mathmate

...and don't forget to add the 2 conditions that correspond to the two factors that you cancelled out.

- pooja195

>:(

- pooja195

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

- mathmate

yes. Please write the answer with the conditions on one line, the conditions are "and" because both have to be satisfied.

- mathmate

If you want to see how it should be done, see the example near the bottom of the page:
http://www.purplemath.com/modules/rtnlmult.htm

- pooja195

Does it have to be written like that? :/

- mathmate

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

- pooja195

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

- mathmate

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

- pooja195

oo right >_< oops

- mathmate

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

- mathmate

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

- pooja195

XD

- mathmate

Got other ones to try?

- pooja195

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

- mathmate

what's "by" O_o

- mathmate

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

- pooja195

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

- mathmate

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

- mathmate

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

- mathmate

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

- pooja195

yes :) xD

- mathmate

Excellent!

- mathmate

What's your answer then?

- mathmate

Bet you this one comes out in the quiz.

- pooja195

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

- mathmate

Another one done, with 100%.

- pooja195

:) MORE!!!:D

- mathmate

yes please! More, more....

- pooja195

You pick :P ;)

- mathmate

I like your examples better, they are less boring.

- mathmate

ok, I'll pick one.

- pooja195

LOL xD

- mathmate

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

- mathmate

There is no grouping to be done in the factoring because the coefficients of the quadratic expressions are all one (1).

- pooja195

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

- mathmate

Wow!

- mathmate

3x-9 = 3(x-3)

- mathmate

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

- mathmate

or you factored all three?

- pooja195

Diffrence of square
\[\huge \frac{ (x+3)(x+3) }{ (x+3)(x-3) } \times \frac{ 3(x-3) }{ (x+3)(x-1) }\]
not sure how to do the next part

- mathmate

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

- mathmate

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

- mathmate

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

- mathmate

ready for another?

- pooja195

yes! :)

- mathmate

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

- pooja195

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

- mathmate

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

- pooja195

:(

- mathmate

That was a great piece of work, involving 4 factorizations,, out of which you got 3 of them perfect, and one was just a switch of the sign! :)

- mathmate

* bottom-right

- pooja195

wait so what should it look like :. ?

- mathmate

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

- mathmate

\(\large \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (x-4)(x+1) }{ (2x-3)(x+1) }\)
Can you finish it?

- mathmate

*four conditions

- pooja195

all of them cancel out! T_T

- mathmate

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

- pooja195

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

- mathmate

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

- mathmate

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

- pooja195

T_T

- pooja195

i wanna move onto something more challenging :)

- mathmate

Like addition and subtration?

- mathmate

*subtraction

- pooja195

ye

- mathmate

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

- mathmate

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

- mathmate

or the common denominator.

- pooja195

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

- pooja195

:/

- mathmate

Excellent!

- mathmate

now we have on denominator (x-4)(x+4) and the other (x+4)(x+4).
Can we find the common denominator?
Example:
common denominator of 6 and 9: 6=2*3, 9=3*3, so common den. = 2*3*3
the common denominator contains all the factor of each of the originals.

- mathmate

*one

- pooja195

x+4

- mathmate

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

- pooja195

;-; idk

- mathmate

(x-1)(x+4)(x+4) contains (x-1)(x+4) which is the first one, and also contains (x+4)(x+4) the second one.
So (x-1)(x+4)(x+4) is the common denominator.
You can also get it by multiplying together the two denominators, but cancelling one of the repetitions.
\((x-4)\color{red}{(x+4)} (x+4)\color{red}{(x+4)}\)
We will keep only one of the two reds because they repeat.
That gives \((x-4)\color{red}{(x+4)} (x+4)\)

- mathmate

Now step 2:
For the first term, we need to multiply top and bottom by (x+2) to get the bottom to be the common denominator, i.e.
\(\large \frac{4\color{blue}{(x+4)}}{(x-4)(x+4)\color{blue}{(x+4)}}\)

- mathmate

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

- mathmate

Since we have a common factor, we can now add:
\(\large \frac{4(x+4)}{(x-4)(x+4)(x+4)}+\frac{3(x-4)}{(x-4)(x+4)(x+4)}\)

- mathmate

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

- mathmate

and that's the answer.

- mathmate

Did you follow all the steps?

- mathmate

If you get stuck, tell me where.

- mathmate

You can also read example 4 of the following link:http://www.cliffsnotes.com/math/algebra/algebra-ii/rational-expressions/adding-and-subtracting-rational-expressions

- mathmate

For the common denominator part, I will rephrase as follows:
You can also get it by multiplying together the two denominators, but cancelling one of the repetitions, if any, \(between\) the denominators.
\([(xâˆ’4)\color{red}{(x+4)}] [(x+4)\color{red}{(x+4)}]\)
We will keep only one of the two reds because they repeat.
That gives \((xâˆ’4)(x+4)\color{red}{(x+4)}\) as the common denominator.

- pooja195

oh ok i get it :)

- mathmate

The next step is to make equivalent fractions by multiplying by appropriate factors to make the denominator the common denominator. This is done for each term to be added/subtracted.

- pooja195

ok =)

- mathmate

Try another one?

- pooja195

No T_T

- mathmate

Do you have anything from your school notes?

- mathmate

Example or exercises?

- pooja195

no o.o

- mathmate

ok, shall we take a break/

- pooja195

mmmmmmmmmmmmmmmm k :)

- pooja195

No lets continue

- mathmate

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

- pooja195

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

- mathmate

Ok, the boring ones!

- mathmate

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

- mathmate

The terms already have a common denominator, so we just add the numerators:
\(\large \frac{4x+3\ \ +\ \ 3x+4}{x^2-16}\)
Can you finish it?

- pooja195

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

- mathmate

Excellent!

- mathmate

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

- mathmate

Like the one before, the two terms already have a common denominator, so just do the subtraction of the numerators and factor, if possible, the difference.

- pooja195

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

- mathmate

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

- pooja195

v_v next

- mathmate

k

- mathmate

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

- pooja195

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

- mathmate

lol, don't get fooled!
The two denminators are actually identical, if you factor the first or expand the second, you will find that they are the same. Just add the numerators and factor if necessary.

- pooja195

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

- mathmate

for me?

- pooja195

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

- mathmate

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

- mathmate

Keep going, you're almost there!

- mathmate

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

- pooja195

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

- mathmate

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

- pooja195

\[x \neq2\]

- mathmate

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

- mathmate

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

- pooja195

got it.

- mathmate

Try the unlike denominators now or later?

- mathmate

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

- mathmate

Please show work!

- pooja195

No lets skip this section i dont like this section

- mathmate

ok, 9=3*3, 12= 3*4
Let's find the LCM:
|dw:1433023297993:dw|
so
\(\large \frac{2}{9}+\frac{7}{12}=\frac{2*4}{9*4}+\frac{7*3}{12*3}=\frac{8}{36}+\frac{21}{36}=\frac{29}{36}\)

- mathmate

@pooja195

- pooja195

I IZ HERE

- mathmate

hold on! have to find the question!

- pooja195

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

- pooja195

mathmate i thin your pm's are lagging

- pooja195

*think

- pooja195

omg

- mathmate

forget the previous one with the wrong operator.
Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8}\div \frac{2x^2-x-3}{x^2-3x-4}\)

- pooja195

omg

- mathmate

First step: flip.

- mathmate

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

- mathmate

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

- pooja195

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

- mathmate

I would do it like that: For the top left, we have -12 and +1.
We know 3*4=12 with a difference of 1 (diff because 12 is negative).
So figure out the right signs, and that will be +4 -3 (to give +1).

- pooja195

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

- mathmate

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

- mathmate

Good job, continue please!

- pooja195

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

- pooja195

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

- mathmate

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

- pooja195

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

- mathmate

so far so good!

- mathmate

Now the finishing touch!

- pooja195

the answer is 1 .-.

- mathmate

right, but not finished yet!

- pooja195

omg no T_T the conditions

- mathmate

The conditions can be obtained by equating each factor cancelled to zero.
Ex. 2x-3\(\ne\)0 means x\(\ne\)3/2, etc.

- mathmate

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

- pooja195

no he doesnt make us list those :/

- mathmate

Never?

- pooja195

nope :/

- mathmate

Even in the quiz? Is it multiple choice?

- mathmate

Ok, do it just for this one,to remind you that they should be there. After that, if they are listed correctly, you don't have to do it until after the quiz!!

- pooja195

;-;

- mathmate

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

- pooja195

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

- mathmate

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

- mathmate

Excellent, thank you! :)

- pooja195

>:(

- mathmate

Ready for another one, simpler, and no conditions!

- pooja195

yes!

- mathmate

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

- mathmate

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

- pooja195

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

- mathmate

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

- mathmate

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

- mathmate

Remember bottom was x^2-2^2

- pooja195

2(2+x)

- mathmate

good, 2+x= x+2
We usually write polynomials in decreasing power. This way it's easier to find like terms.

- mathmate

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

- pooja195

answer is 1

- mathmate

I think it's 2. Can you check?

- pooja195

yh its 2 ;o

- mathmate

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

- pooja195

None of the above. xD

- mathmate

ok, it's getting late!

- pooja195

LOL ok we can continue tommzzz

- pooja195

*tommrow

- mathmate

ok!

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