At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Yep. Do you know direct and inverse variation?
no ._.
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh ok :)
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.
ok with direct?
yes :)
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|
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.
The sum of two rational expressions is also a rational expression, just like the sum of two fractions is still a fraction.
Even with one single rational expression, we can simplify , for example, what would be \(\large \frac{(4x+3)(x-1)}{(x-1)}\) ?
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
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: \(\large \frac{4x^2-x-3}{x-1}\) ?
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| \[(4x^2-4x)+(3x-3)\] \[4x(x-1)+3(x-1)\] \[(4x+3)(x-1)\]
Excellent! Now you replace the numerator of the rational expression with what you've got, what do you get? \(\large \frac{4x^2-x-3}{x-1}= ?\)
\[\frac{ (4x+3)(x-1) }{ (x-1) }\] (x-1) cancel out. Final answer: \[ \huge 4x+3\]
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).
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.
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.
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
Are we good so far (except you don't like the condition! :) )
yes :)
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
ok so far?
yes.
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
ok so far?
yes :)
Now for adding and subtracting with like denominators, it would be just like in fractions. Should I give an example anyway?
Can we just skip those :P i know those
What about with unlike denominators, are you ok with those?
no...
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?
yes
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?
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
We need the product of the two denominators, (x-2)(x+4), so that we can make equivalent expressions. \(\large \frac{x+1}{x-2}-\frac{x-2}{x+4}\) =\(\large \frac{(x+1)(x+4)}{(x-2)(x+4)}-\frac{(x-2)(x-2)}{(x+4)(x-2)}\)
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?
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).
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!
@pooja195 Time to work!
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
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
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
Remember, when the product (-6) is negative, the two numbers you're looking for have different signs. Does that help?
|dw:1432934285205:dw| Which pair works?
3 -2 .-.
(x+3)(x-2)
Exactly. Are you able to complete the answer now?
\[\frac{ (x-2)}{ (x+4) }\]
Almost, just a transcription error (should be x-4 at the bottom) AND missing very important information... Do you know what's missing?
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 :)
Just have to remember that whenever you cancel a term, it comes with a condition. Like when you buy candies, you always pay! :)
Want another one similar?
no :(
T_T !
ok fine another one
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 \(\large \frac{x^2+x-6}{x^2-2x-15}\)
|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!!! >:(
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.
:)
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)
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
Ahhh >:( T_T
Now try this: simplify (2x+5/2)(x+3)/(2x+5/2)
No no more of these T_T
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!
x+3 ;-;
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
Can we please skip these T_T
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.
Simplify (2x+3)(x-1)/(x-1)
2x+3 x/= 1??????
yes, put if, for, for all, or whenever... to indicate that it is a condition. so 2x+3 for all x\(\ne\) 1
Are you ok now?
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?
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) \[\huge \frac{ x-2 }{ x-4 }\neq 3\]
That's the idea! We write the condition separate though, \(\large \frac{ x-2 }{ x-4 }\ for\ all\ x\neq 3\)
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! :|
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
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) \[\frac{ (x-3)(5x+3) }{ (5x+6)(x-2) }\neq-2\]
you cancelled the x-2 at the bottom, so x-2\(\ne\)0, or x\(\ne\)2 The answer will then read: \(\frac{ (x-3)(5x+3) }{ (5x+6)} \ for\ all\ x\neq2\)
>_<
Simplify \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x-6}\)
You'll need to invert the second term to change division to multiplication: =\(\large \frac{x-3}{5x+6}\times \frac{5x-6}{x-2}\)
oops the question is wrong. It is meant to have 5x+6 top and bottom! \(\large \frac{x-3}{5x+6}\div \frac{x-2}{5x+6}\) =\(\large \frac{x-3}{5x+6}\times \frac{5x+6}{x-2}\)
O_o
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
ok i kinda get it
@pooja195 we were at 11.4 are you ready?
ye
It seems better now. Are you good with multiplication and division of rationals?
kind of :/
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.
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?
That's ok, but I would prefer to put the second term upside down, so we will always working with multiplication.
* 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?
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.
\(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)
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?
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.
So you can go ahead and factor, as though they are two separate terms. We can combine them after cancelling.
huh :?
I mean you can proceed with this: \(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x^2-9x }\)
How would i write out the factors?
The bottom one on the right should take out 3x as well, so you put \(\large \frac{ x^2+2x-15 }{ 3x(x+2) }\times \frac{ x^2+7x+10 }{ 3x(x-3) }\)
For the ones in the numerator, you can put them where they are now.
If you want, we can factor the left numerator together.
\(\large x^2+2x-15\) =\(\large (x- \ )(x+ \ )\)
ok |dw:1433012589434:dw| |dw:1433012632806:dw| \[\frac{ (x+5)(x-3) }{ 3x(x+2) } \times \frac{ (x+5)(x+2) }{ 3x(x-3) }\]
Good job! Speed of a bullet!
can you now post me the factors that you can cancel, and write down the condition associated with each one?
like: (x-3) means x\(\ne\)3
You can write all the factors on top as one single numerator, and similarly for the denominators. Looks like this:
\[\frac{ (x+5)^2 }{ 3x }\]
Yes, but there are two "3x" , so you write 9x^2 at the bottom.
...and don't forget to add the 2 conditions that correspond to the two factors that you cancelled out.
>:(
\[x \neq-3~~~~~~~x \neq2\]
yes. Please write the answer with the conditions on one line, the conditions are "and" because both have to be satisfied.
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
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 \(\large \frac{x^2+6x+9}{x^2-9}.\frac{3x-9}{x^2+2x-3}\)
There is no grouping to be done in the factoring because the coefficients of the quadratic expressions are all one (1).
\[\frac{ (x+3)(x+3) }{ (x+3)(x-3) }\times \frac{ 3x-9 }{x+3)(x-1) }\]
Wow!
3x-9 = 3(x-3)
Did you use the perfect square and diff. of two squares?
or you factored all three?
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
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.
:(
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! :)
* bottom-right
wait so what should it look like :. ?
The answer should be very simple, almost the simplest possible, plus three conditions.
\(\large \frac{ (2x-3)(x+2) }{ (x-4)(x+2) }\times \frac{ (x-4)(x+1) }{ (2x-3)(x+1) }\) Can you finish it?
*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!
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.
*one
x+4
x+4 does not contain x-4, and x+4 twice. Hint: the common denominator here has three factors.
;-; idk
(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 \((x-4)\color{red}{(x+4)} (x+4)\)
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. \(\large \frac{4\color{blue}{(x+4)}}{(x-4)(x+4)\color{blue}{(x+4)}}\)
Similarly for the second term: \(\large \frac{3\color{blue}{(x-4)}}{\color{blue}{(x-4)}(x+4)(x+4)}\)
Since we have a common factor, we can now add: \(\large \frac{4(x+4)}{(x-4)(x+4)(x+4)}+\frac{3(x-4)}{(x-4)(x+4)(x+4)}\)
= \(\large \frac{4x+16+3x-12}{(x-4)(x+4)(x+4)}\) = \(\large \frac{7x+4}{(x-4)(x+4)(x+4)}\)
and that's the answer.
Did you follow all the steps?
If you get stuck, tell me where.
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
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.
oh ok i get it :)
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.
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 \(\large \frac{4x+3}{x^2-16}+\frac{3x+4}{x^2-16}\)
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?
\[\frac{7\left(x+1\right)}{\left(x+4\right)\left(x-4\right)} \]
Excellent!
Simplify \(\large \frac{x^2-3x+2}{2x-4}-\frac{x-1}{2x-4}\)
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.
\[\frac{\left(x-3\right)\left(x-1\right)}{2\left(x-2\right)} \]
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
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.
\[\huge\frac{ 2x+3}{ (x+2)(x+2)}+\frac{ (x+1)(x+1) }{ (x+2)(x+2) }\]
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
ok, 9=3*3, 12= 3*4 Let's find the LCM: |dw:1433023297993:dw| so \(\large \frac{2}{9}+\frac{7}{12}=\frac{2*4}{9*4}+\frac{7*3}{12*3}=\frac{8}{36}+\frac{21}{36}=\frac{29}{36}\)
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
forget the previous one with the wrong operator. Simplify \(\large \frac{2x^2+x-6}{x^2-2x-8}\div \frac{2x^2-x-3}{x^2-3x-4}\)
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!
\[\huge \frac{ 2x^2+x-6 }{ x^2-2x-8 }\times \frac{ x^2-3x-4 }{ 2x^2-x-3 }\]
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).
\[huge(x+4)(x-3)\] \[\huge(2x^2+4x)(-3x-6)\] \[\huge2x(x+2)~~~~~-3(x+2)\]
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
The conditions can be obtained by equating each factor cancelled to zero. Ex. 2x-3\(\ne\)0 means x\(\ne\)3/2, etc.
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?
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!!
;-;
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)
good, 2+x= x+2 We usually write polynomials in decreasing power. This way it's easier to find like terms.
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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question