pooja195
  • pooja195
@mathmate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathmate
  • mathmate
Yep. Do you know direct and inverse variation?
pooja195
  • pooja195
no ._.
mathmate
  • 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?

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More answers

pooja195
  • pooja195
oh ok :)
mathmate
  • 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
  • mathmate
ok with direct?
pooja195
  • pooja195
yes :)
mathmate
  • 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
  • mathmate
ok for inverse variation?
pooja195
  • pooja195
Yes i get that :)
mathmate
  • mathmate
Can we move on the rational expressions?
mathmate
  • mathmate
* to
pooja195
  • pooja195
yes
mathmate
  • mathmate
A rational expression is a polynomial divided by another polynomial.
mathmate
  • mathmate
Example, (5x+3)/(2x+1) is a rational expression.
mathmate
  • mathmate
The sum of two rational expressions is also a rational expression, just like the sum of two fractions is still a fraction.
mathmate
  • mathmate
Even with one single rational expression, we can simplify , for example, what would be \(\large \frac{(4x+3)(x-1)}{(x-1)}\) ?
pooja195
  • pooja195
i just foil right?
mathmate
  • mathmate
Before you do foil, did you notice anything?
pooja195
  • pooja195
we can cancel
mathmate
  • mathmate
Yes, only as long as (x-1) does not equal zero!
pooja195
  • pooja195
Yes :)
mathmate
  • mathmate
So we say: \(\large \frac{(4x+3)(x-1)}{(x-1)}=4x+3\) if x\(\ne\)1
mathmate
  • mathmate
ok with that?
pooja195
  • pooja195
Yes :D
mathmate
  • 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
  • mathmate
We cannot tell off-hand if there is any common factor, so what do we do?
pooja195
  • pooja195
We use the wheel thingy
mathmate
  • mathmate
Good, or "factorize" is the other word for the wheel thingy! Can you do that, please?
pooja195
  • pooja195
|dw:1432867194095:dw| \[(4x^2-4x)+(3x-3)\] \[4x(x-1)+3(x-1)\] \[(4x+3)(x-1)\]
mathmate
  • 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
  • pooja195
\[\frac{ (4x+3)(x-1) }{ (x-1) }\] (x-1) cancel out. Final answer: \[ \huge 4x+3\]
mathmate
  • 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
  • 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
  • mathmate
So the answer is again 4x+3 with x\(\ne\)1.
pooja195
  • pooja195
I dont like that :/
mathmate
  • mathmate
the condition?
pooja195
  • pooja195
Yep
mathmate
  • mathmate
Yes, it complicates life a little, but that's the only difference from fractions.
mathmate
  • 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
  • mathmate
Are we good so far (except you don't like the condition! :) )
pooja195
  • pooja195
yes :)
mathmate
  • 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
  • mathmate
ok so far?
pooja195
  • pooja195
yes.
mathmate
  • 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
  • mathmate
ok so far?
pooja195
  • pooja195
yes :)
mathmate
  • mathmate
Now for adding and subtracting with like denominators, it would be just like in fractions. Should I give an example anyway?
pooja195
  • pooja195
Can we just skip those :P i know those
mathmate
  • mathmate
What about with unlike denominators, are you ok with those?
pooja195
  • pooja195
no...
mathmate
  • 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
  • pooja195
yes
mathmate
  • 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
  • pooja195
yes
mathmate
  • mathmate
Now for rational exprssions: \(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\)
mathmate
  • mathmate
What would be the common denominator? (do not FOIL, otherwise we need to factorize afterwards.)
pooja195
  • pooja195
x+4
mathmate
  • 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
  • mathmate
ok?
mathmate
  • mathmate
To this we have to add the conditions that x-2\(\ne\)0 and x+4\(\ne\)0
mathmate
  • mathmate
ok so far in making equivalent fractions?
mathmate
  • 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
  • pooja195
im so confused and lost T_T
mathmate
  • mathmate
@pooja195 xD
pooja195
  • pooja195
NO!! T_T
mathmate
  • mathmate
Yesterday we went through 11.2 to 11.6. Now it's time to do some practice!
pooja195
  • pooja195
next class
mathmate
  • mathmate
:( your way out!
mathmate
  • mathmate
Just kidding, sure, later!
mathmate
  • mathmate
@pooja195 Time to work!
pooja195
  • pooja195
O_O
pooja195
  • pooja195
Just remebered i have a thing to do.... oops cant work :P
mathmate
  • mathmate
Ok, I'll grab you later, ok?
pooja195
  • pooja195
BWAHAHAHA XDDD :3 yr so innocent xS
pooja195
  • pooja195
im free :# we can math :P
mathmate
  • mathmate
ok! I was O_o
mathmate
  • mathmate
R U good with 11.3?
pooja195
  • pooja195
:/
mathmate
  • mathmate
Want a practice to make sure?
pooja195
  • pooja195
yes
mathmate
  • mathmate
BTW, were the correct answers in the quiz all between 11.2 and 11.3?
pooja195
  • pooja195
no :/
mathmate
  • mathmate
So you did get some in other sections, that's good!
pooja195
  • pooja195
ummm no
mathmate
  • 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
  • pooja195
i thought i did great on it but i ended up failing
pooja195
  • pooja195
;-; can we start with something easier :(
mathmate
  • mathmate
Ok, simplify \(\large f(x)=\frac{x^2+x-6}{(x+3)(x-4)}\) :)
pooja195
  • pooja195
|dw:1432933988455:dw| IDK T_T
mathmate
  • mathmate
Remember, when the product (-6) is negative, the two numbers you're looking for have different signs. Does that help?
mathmate
  • mathmate
|dw:1432934285205:dw| Which pair works?
pooja195
  • pooja195
3 -2 .-.
pooja195
  • pooja195
(x+3)(x-2)
mathmate
  • mathmate
Exactly. Are you able to complete the answer now?
pooja195
  • pooja195
\[\frac{ (x-2)}{ (x+4) }\]
mathmate
  • 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
  • pooja195
no
mathmate
  • mathmate
The condition that (x+3)\(\ne\)0. Teachers are waiting to jump on that! :(
mathmate
  • mathmate
So the complete answer is (x-2)/(x-4) with x\(\ne\) -3
mathmate
  • mathmate
ok so far?
pooja195
  • pooja195
I hate that
pooja195
  • pooja195
ye the rest is fine :)
mathmate
  • 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
  • mathmate
Want another one similar?
pooja195
  • pooja195
no :(
mathmate
  • mathmate
T_T !
pooja195
  • pooja195
ok fine another one
mathmate
  • 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
  • pooja195
|dw:1432935523037:dw| \[\frac{(x-2)(x+3) }{ (x-5)(5+3)}\] Final answer \[\frac{ (x-2) }{ (x+5) }\]
mathmate
  • mathmate
oh...oh
pooja195
  • pooja195
O_O ?
mathmate
  • mathmate
Remember the last problem? transcription and something missing???
pooja195
  • pooja195
T_T i dont know this!!! >:(
mathmate
  • 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
  • mathmate
@pooja195
pooja195
  • pooja195
:)
mathmate
  • mathmate
We were on 11.3
mathmate
  • mathmate
All you have to remember is if you get candies, you pay!
mathmate
  • mathmate
candies ~ cancel
pooja195
  • pooja195
:( i hate that
mathmate
  • mathmate
Sorry, we have to stick to rules! Now can you simplify (x-2)(x+4)/(x+4) ?
pooja195
  • pooja195
(x-2)
mathmate
  • mathmate
You cancelled (x+4) right?
pooja195
  • pooja195
yea....
mathmate
  • mathmate
So we write (x-2).....
pooja195
  • pooja195
thats what i did .-.
mathmate
  • mathmate
You were close: (x-2) for x\(\ne\)2 is the complete answer.
mathmate
  • mathmate
* for x\(\ne\) -4
pooja195
  • pooja195
>:(
mathmate
  • mathmate
ok, let's try another: simplify (x-4)(x+2)/(x-4)
pooja195
  • pooja195
(x+2)
mathmate
  • 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
  • pooja195
Ahhh >:( T_T
mathmate
  • mathmate
Now try this: simplify (2x+5/2)(x+3)/(2x+5/2)
pooja195
  • pooja195
No no more of these T_T
mathmate
  • 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
  • pooja195
x+3 ;-;
mathmate
  • 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
  • pooja195
Can we please skip these T_T
mathmate
  • 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
  • mathmate
Simplify (2x+3)(x-1)/(x-1)
pooja195
  • pooja195
2x+3 x/= 1??????
mathmate
  • 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
  • mathmate
Are you ok now?
mathmate
  • 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
  • pooja195
Yes
mathmate
  • mathmate
Ready for 11.4?
pooja195
  • pooja195
Yes
mathmate
  • mathmate
@pooja195
pooja195
  • pooja195
I iz here
mathmate
  • mathmate
11.4 is on rational expressions and not radical expression, I suppose. Can you confirm?
pooja195
  • pooja195
yea lol rational is right
mathmate
  • mathmate
ok, try this: simplify \(\large \frac{x-2}{x+3}\times\frac{x+3}{x-4}\)
pooja195
  • pooja195
(x-2)(x+3)/(x+3)(x-4) \[\huge \frac{ x-2 }{ x-4 }\neq 3\]
mathmate
  • mathmate
That's the idea! We write the condition separate though, \(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq 3\)
mathmate
  • mathmate
correction: \(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq -3\) because we don't want x+3=0
mathmate
  • mathmate
Now, try Simplify \(\large \frac{x-2}{5x+6}\times\frac{2x-3}{x-2}\)
pooja195
  • pooja195
(x-2)(x-2)/(5x+6)(2x-3) T_T Lets do a diffrent one! :|
mathmate
  • 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
  • mathmate
ok try this: Simplify \(\large \frac{x^2-5x+6}{5x+6}\times\frac{5x+3}{x-2}\)
pooja195
  • pooja195
(x-3)(x-2)(5x+3)/(5x+6)(x-2) \[\frac{ (x-3)(5x+3) }{ (5x+6)(x-2) }\neq-2\]
mathmate
  • 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
  • pooja195
>_<
mathmate
  • mathmate
Simplify \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x-6}\)
mathmate
  • 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
  • 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
  • pooja195
O_o
mathmate
  • 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
  • pooja195
ok i kinda get it
mathmate
  • mathmate
@pooja195 we were at 11.4 are you ready?
pooja195
  • pooja195
ye
mathmate
  • mathmate
It seems better now. Are you good with multiplication and division of rationals?
pooja195
  • pooja195
kind of :/
mathmate
  • 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
  • pooja195
yeah lol my hoe work ill put up a problem from that : )
mathmate
  • mathmate
Sure! It will be less boring.
pooja195
  • pooja195
xD \[\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ 3x^2-9x }{ x^2+7x+10 }\]
mathmate
  • mathmate
So what would be your first step?
pooja195
  • pooja195
factor?
mathmate
  • mathmate
That's ok, but I would prefer to put the second term upside down, so we will always working with multiplication.
mathmate
  • mathmate
* always be...
pooja195
  • pooja195
ooo right flip
mathmate
  • mathmate
\(\frac{ x^2+2x-15 }{ 3x^2+6x }\div \frac{ x^2+7x+10 }{ 3x^2-9x }\)
mathmate
  • mathmate
Oops too small, but I can still read it.
pooja195
  • pooja195
lol ok so now we factor right?
mathmate
  • 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
  • mathmate
\(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)
pooja195
  • 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
  • mathmate
Actually on second look, would the original question be a multiplication instead of a division?
mathmate
  • mathmate
You can start with writing them separately in two separate numerators. After that, we cancel.
mathmate
  • mathmate
But it seems that there is not much to cancel unless the original question was a multiplication.
pooja195
  • pooja195
its division :/
mathmate
  • mathmate
Sorry, I take it back. There is at least one that we can cancel.
mathmate
  • mathmate
So you can go ahead and factor, as though they are two separate terms. We can combine them after cancelling.
pooja195
  • pooja195
huh :?
mathmate
  • 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
  • pooja195
How would i write out the factors?
mathmate
  • 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
  • mathmate
For the ones in the numerator, you can put them where they are now.
mathmate
  • mathmate
If you want, we can factor the left numerator together.
mathmate
  • mathmate
\(\large x^2+2x-15\) =\(\large (x- \ )(x+ \ )\)
pooja195
  • 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
  • mathmate
Good job! Speed of a bullet!
mathmate
  • mathmate
can you now post me the factors that you can cancel, and write down the condition associated with each one?
mathmate
  • mathmate
like: (x-3) means x\(\ne\)3
mathmate
  • mathmate
You can write all the factors on top as one single numerator, and similarly for the denominators. Looks like this:
pooja195
  • pooja195
\[\frac{ (x+5)^2 }{ 3x }\]
mathmate
  • mathmate
Yes, but there are two "3x" , so you write 9x^2 at the bottom.
mathmate
  • mathmate
...and don't forget to add the 2 conditions that correspond to the two factors that you cancelled out.
pooja195
  • pooja195
>:(
pooja195
  • pooja195
\[x \neq-3~~~~~~~x \neq2\]
mathmate
  • mathmate
yes. Please write the answer with the conditions on one line, the conditions are "and" because both have to be satisfied.
mathmate
  • 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
  • pooja195
Does it have to be written like that? :/
mathmate
  • mathmate
It would be the simplest, simpler than what I would have done.
pooja195
  • pooja195
\[\frac{ (x+5)^2 }{ 3x },x \neq2,x \neq-3\]
mathmate
  • 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
  • pooja195
oo right >_< oops
mathmate
  • mathmate
\(\large\frac{ (x+5)^2 }{ 9x^2 },x \neq2,x \neq-3\)
mathmate
  • mathmate
Excellent, you got one done, and it wasn't a simple one like my (boring) ones.
pooja195
  • pooja195
XD
mathmate
  • mathmate
Got other ones to try?
pooja195
  • pooja195
\[\huge \frac{ x^2-64 }{ 3x^3 }by~(8-8)\]
mathmate
  • mathmate
what's "by" O_o
mathmate
  • mathmate
Sure it's (8-8) ? (equals zero!) O_o!
pooja195
  • pooja195
\[\huge \frac{ (x+8)(x-8) }{ 3x^3}\div \frac{ 1 }{ (x-8) }\]
mathmate
  • mathmate
Ok! I think you did the division twice (you flipped, and you kept the division sign!)
mathmate
  • mathmate
Is it \(\huge \frac{ (x+8)(x-8) }{ 3x^3}\times \frac{ 1 }{ (x-8) }\)
mathmate
  • mathmate
And you factored the top right-away, ;) ?
pooja195
  • pooja195
yes :) xD
mathmate
  • mathmate
Excellent!
mathmate
  • mathmate
What's your answer then?
mathmate
  • mathmate
Bet you this one comes out in the quiz.
pooja195
  • pooja195
\[\huge\frac{ 1(x+8) }{ 3x^3 },x \neq -8\]
mathmate
  • mathmate
Another one done, with 100%.
pooja195
  • pooja195
:) MORE!!!:D
mathmate
  • mathmate
yes please! More, more....
pooja195
  • pooja195
You pick :P ;)
mathmate
  • mathmate
I like your examples better, they are less boring.
mathmate
  • mathmate
ok, I'll pick one.
pooja195
  • pooja195
LOL xD
mathmate
  • mathmate
Simplify \(\large \frac{x^2+6x+9}{x^2-9}.\frac{3x-9}{x^2+2x-3}\)
mathmate
  • mathmate
There is no grouping to be done in the factoring because the coefficients of the quadratic expressions are all one (1).
pooja195
  • pooja195
\[\frac{ (x+3)(x+3) }{ (x+3)(x-3) }\times \frac{ 3x-9 }{x+3)(x-1) }\]
mathmate
  • mathmate
Wow!
mathmate
  • mathmate
3x-9 = 3(x-3)
mathmate
  • mathmate
Did you use the perfect square and diff. of two squares?
mathmate
  • mathmate
or you factored all three?
pooja195
  • 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
  • mathmate
Excellent. Now you cancel, and note the conditions as you cancel.
mathmate
  • mathmate
(x-3) and (x+3)^2 will be cancelled top and bottom, leaving \(\large \frac{3}{x-1}\), x\(\ne\)-3,1
mathmate
  • mathmate
* \(\large \frac{3}{x-1}\), x\(\ne\)-3,3 You don't have to worry about x+3 twice.
mathmate
  • mathmate
ready for another?
pooja195
  • pooja195
yes! :)
mathmate
  • mathmate
Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8} \div \frac{2x^2-x-3}{x^2-3x-4}\)
pooja195
  • pooja195
\[\frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (2x-3)(x+1) }{ (x+4)(x-1) }\]
mathmate
  • mathmate
Great, just you need to flip, and I think bottom left should read (x-4)(x+1) before flipping.
pooja195
  • pooja195
:(
mathmate
  • 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
  • mathmate
* bottom-right
pooja195
  • pooja195
wait so what should it look like :. ?
mathmate
  • mathmate
The answer should be very simple, almost the simplest possible, plus three conditions.
mathmate
  • 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
  • mathmate
*four conditions
pooja195
  • pooja195
all of them cancel out! T_T
mathmate
  • mathmate
That's good, so what's the answer? (it's not zero).
pooja195
  • pooja195
\[\huge x~\neq-4~~~x~\neq2~~~~~x~\neq1~~~~~????\]
mathmate
  • 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
  • mathmate
Continue with these "easy" ones, or some addition?
pooja195
  • pooja195
T_T
pooja195
  • pooja195
i wanna move onto something more challenging :)
mathmate
  • mathmate
Like addition and subtration?
mathmate
  • mathmate
*subtraction
pooja195
  • pooja195
ye
mathmate
  • mathmate
ok, Simplify \(\large \frac{4}{x^2-16}+\frac{3}{x^2+8x+16}\)
mathmate
  • mathmate
Step 1: factorize, then we can find the lowest common multiple (LCM)
mathmate
  • mathmate
or the common denominator.
pooja195
  • pooja195
\[\huge \frac{ 4 }{ (x+4)(x-4) }+\frac{ 3 }{(x+4)(x+4) }\]
pooja195
  • pooja195
:/
mathmate
  • mathmate
Excellent!
mathmate
  • 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
  • mathmate
*one
pooja195
  • pooja195
x+4
mathmate
  • mathmate
x+4 does not contain x-4, and x+4 twice. Hint: the common denominator here has three factors.
pooja195
  • pooja195
;-; idk
mathmate
  • 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
  • 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
  • mathmate
Similarly for the second term: \(\large \frac{3\color{blue}{(x-4)}}{\color{blue}{(x-4)}(x+4)(x+4)}\)
mathmate
  • 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
  • mathmate
= \(\large \frac{4x+16+3x-12}{(x-4)(x+4)(x+4)}\) = \(\large \frac{7x+4}{(x-4)(x+4)(x+4)}\)
mathmate
  • mathmate
and that's the answer.
mathmate
  • mathmate
Did you follow all the steps?
mathmate
  • mathmate
If you get stuck, tell me where.
mathmate
  • 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
  • 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. \([(x−4)\color{red}{(x+4)}] [(x+4)\color{red}{(x+4)}]\) We will keep only one of the two reds because they repeat. That gives \((x−4)(x+4)\color{red}{(x+4)}\) as the common denominator.
pooja195
  • pooja195
oh ok i get it :)
mathmate
  • 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
  • pooja195
ok =)
mathmate
  • mathmate
Try another one?
pooja195
  • pooja195
No T_T
mathmate
  • mathmate
Do you have anything from your school notes?
mathmate
  • mathmate
Example or exercises?
pooja195
  • pooja195
no o.o
mathmate
  • mathmate
ok, shall we take a break/
pooja195
  • pooja195
mmmmmmmmmmmmmmmm k :)
pooja195
  • pooja195
No lets continue
mathmate
  • mathmate
Do you have some examples for the easy ones (like denominators)?
pooja195
  • pooja195
No. I'd prefer you give the problems... :/
mathmate
  • mathmate
Ok, the boring ones!
mathmate
  • mathmate
Simplify \(\large \frac{4x+3}{x^2-16}+\frac{3x+4}{x^2-16}\)
mathmate
  • 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
  • pooja195
\[\frac{7\left(x+1\right)}{\left(x+4\right)\left(x-4\right)} \]
mathmate
  • mathmate
Excellent!
mathmate
  • mathmate
Simplify \(\large \frac{x^2-3x+2}{2x-4}-\frac{x-1}{2x-4}\)
mathmate
  • 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
  • pooja195
\[\frac{\left(x-3\right)\left(x-1\right)}{2\left(x-2\right)} \]
mathmate
  • mathmate
Oh yes, I thought the difference is not factorable! lol Great job!
pooja195
  • pooja195
v_v next
mathmate
  • mathmate
k
mathmate
  • mathmate
\(\large \frac{2x+3}{x^2+4x+4} +\frac{x^2+2x+1}{(x+2)^2}\)
pooja195
  • pooja195
;-; not this one! Lets skip it! :D
mathmate
  • 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
  • pooja195
\[\huge\frac{ 2x+3}{ (x+2)(x+2)}+\frac{ (x+1)(x+1) }{ (x+2)(x+2) }\]
mathmate
  • mathmate
for me?
pooja195
  • pooja195
\[\huge\frac{ 2x+3(x+1)^2}{ (x+2)(x+2) }\]
mathmate
  • mathmate
Watch out: \(\huge\frac{ 2x+3\color{red}{+}(x+1)^2}{ (x+2)(x+2) }\)
mathmate
  • mathmate
Keep going, you're almost there!
mathmate
  • mathmate
You need to expand the (x+1)^2 using FOIL or identities and then add.
pooja195
  • 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
  • mathmate
Great! So after cancelling you have a numeric answer with a single condition:
pooja195
  • pooja195
\[x \neq2\]
mathmate
  • mathmate
The answer is 1 (5*5/(5*5)=1) and the condition is x+2\(\ne\)0, or x\(\ne\)-2
mathmate
  • mathmate
Remember we don't want what we cancelled to be zero.
pooja195
  • pooja195
got it.
mathmate
  • mathmate
Try the unlike denominators now or later?
mathmate
  • mathmate
I'll start with this: Add \(\large \frac{2}{9}+\frac{7}{12}\)
mathmate
  • mathmate
Please show work!
pooja195
  • pooja195
No lets skip this section i dont like this section
mathmate
  • 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
  • mathmate
@pooja195
pooja195
  • pooja195
I IZ HERE
mathmate
  • mathmate
hold on! have to find the question!
pooja195
  • pooja195
ugh no i put everything away ;-; lets just use yours :(
pooja195
  • pooja195
mathmate i thin your pm's are lagging
pooja195
  • pooja195
*think
pooja195
  • pooja195
omg
mathmate
  • 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
  • pooja195
omg
mathmate
  • mathmate
First step: flip.
mathmate
  • mathmate
\(\large \frac{2x^2+x-6}{x^2-2x-8}\times \frac{x^2-3x-4}{2x^2-x-3}\)
mathmate
  • mathmate
Next, factor, one piece at a time, and show intermediate steps, please!
pooja195
  • pooja195
\[\huge \frac{ 2x^2+x-6 }{ x^2-2x-8 }\times \frac{ x^2-3x-4 }{ 2x^2-x-3 }\]
mathmate
  • 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
  • pooja195
\[huge(x+4)(x-3)\] \[\huge(2x^2+4x)(-3x-6)\] \[\huge2x(x+2)~~~~~-3(x+2)\]
mathmate
  • mathmate
Then proceed with grouping: 2x^2+4x -3x-6 = 2x(x+2) -3(x+2) = (2x-3)(x+2)
mathmate
  • mathmate
Good job, continue please!
pooja195
  • pooja195
[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]
pooja195
  • pooja195
\[(2x^2+2x)(-3x-3)\] \[2x(x+1)~~~~~~~-3(x+1)\] \[(2x-3)(x+1)\]
mathmate
  • mathmate
You mean \[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ 2x^2-x-3 }\]
pooja195
  • pooja195
\[[\huge \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{(x-4)(x+1) }{ (2x-3)(x+1) }\] \]
mathmate
  • mathmate
so far so good!
mathmate
  • mathmate
Now the finishing touch!
pooja195
  • pooja195
the answer is 1 .-.
mathmate
  • mathmate
right, but not finished yet!
pooja195
  • pooja195
omg no T_T the conditions
mathmate
  • 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
  • mathmate
Didn't your teacher ask you to specify the conditions, or is it just I?
pooja195
  • pooja195
no he doesnt make us list those :/
mathmate
  • mathmate
Never?
pooja195
  • pooja195
nope :/
mathmate
  • mathmate
Even in the quiz? Is it multiple choice?
mathmate
  • 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
  • pooja195
;-;
mathmate
  • mathmate
So the four conditions are: 2x-3\(\ne\)0 x+2\(\ne\)0 x-4\(\ne\)0 x+1\(\ne\)0
pooja195
  • pooja195
\[x \neq 3/2~~~~~x \neq-2~~~x \neq4~~x \neq-1\]
mathmate
  • mathmate
So you can write them together as: x\(\ne\)-2,-1,3/2,4
mathmate
  • mathmate
Excellent, thank you! :)
pooja195
  • pooja195
>:(
mathmate
  • mathmate
Ready for another one, simpler, and no conditions!
pooja195
  • pooja195
yes!
mathmate
  • mathmate
\(\large \frac{4+2x}{x^2-4}.\frac{x^2-4x+4}{x-2}\)
mathmate
  • mathmate
Hint: x^2-4 is the same as x^2-2^2, difference of 2 squares.
pooja195
  • pooja195
\[\frac{ 4+2x }{ (x+4)(x-4) }\times \frac{ (x-2)(x-2) }{(x-2) }\]
mathmate
  • mathmate
and x^2-4x+4 is a perfect square, if you recall!
mathmate
  • mathmate
You can do better with the first expression (on the left, both top and bottom.
mathmate
  • mathmate
Remember bottom was x^2-2^2
pooja195
  • pooja195
2(2+x)
mathmate
  • mathmate
good, 2+x= x+2 We usually write polynomials in decreasing power. This way it's easier to find like terms.
mathmate
  • mathmate
The bottom is (x+2)(x-2)
pooja195
  • pooja195
answer is 1
mathmate
  • mathmate
I think it's 2. Can you check?
pooja195
  • pooja195
yh its 2 ;o
mathmate
  • mathmate
Good. Want to do rational addition/subtraction (more difficult) or rational equations?
pooja195
  • pooja195
None of the above. xD
mathmate
  • mathmate
ok, it's getting late!
pooja195
  • pooja195
LOL ok we can continue tommzzz
pooja195
  • pooja195
*tommrow
mathmate
  • mathmate
ok!

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