Question

[(1/(x^2)-1/Y^2)]/[(1/x^2+2/xy+1/y^2)]
plz help! youd be awsome if ya did! (tell me how you get then answer as well, thanks!)

Answer 1

What are your instructions for this problem?

Answer 2

"simplify complex functions"

Answer 3

OK, I started by multiplying the top and bottom of the complex fraction by the LCD, which is x^2y^2 in this case

Answer 4

\[\frac{\frac{1}{x^2}-\frac{1}{2}}{\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}}\]

Answer 5

arya has it, just wanted to type it.

Answer 6

Which came out to (y^2-x^2)/(y^2+2xy+x^2)

Answer 7

Sorry, haven't figured out how to type all of the nice equations yet

Answer 8

there is a gimmick here. the denominator is \[(\frac{1}{x}+\frac{1}{y})^2\]

Answer 9

and the numerator is \[(\frac{1}{x}+\frac{1}{y})(\frac{1}{x}-\frac{1}{y})\]

Answer 10

so you can cancel.

Answer 11

can you explain how multiplying the LCD with the second fraction on the bottom came out to be 2xy?

Answer 12

nevermind i got it

Answer 13

\[(2/xy)timesx ^{2}y ^{2}=2x ^{2}y ^{2}/xy=2\]

Answer 14

idk what to do after (y^2-x^2)/(y^2+2xy+x^2)

Answer 15

Ignore my response, it didn't come out quite right

Answer 16

Factor

Answer 17

\[\frac{a^2-b^2}{a+2ab+b^2}=\frac{(a+b)(a-b)}{(a+b)(a+b)}=\frac{a-b}{a+b}\]

Answer 18

y^2-x^2=(y-x)(y+x)
y^2+2xy+x^2=(y+x)(y+x)

Answer 19

\[\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}}\]
multiply top an bottom by \[xy\]

Answer 20

Then we can cancel one of the (y+x)

Answer 21

oh ok thanks so much! think you could help me with [a-b]/[a^-1=b^-1]

Answer 22

Sure

Answer 23

imean [a-b]/[a^-1-b^-1]

Answer 24

OK, a^-1=1/a and b^-1=1/b

Answer 25

Does that part make sense?

Answer 26

no? so ur saying you just move the -1 to the top and not the a along with it?

Answer 27

Not quite, lets see if it makes more sense when I type out the whole equation. Hang on

Answer 28

(a-b)/(a^-1-b^-1) = (a-b)/[(1/a)-(1/b)]

Answer 29

Does that make any more sense or no?

Answer 30

oh ok ya that makes sense, so you dont move the a with the 1 im guessing

Answer 31

\[\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x y}}=1-\frac{2 x}{x+y} \]
Both equation sides evaluated at x=11 and y=17 yield the value 3/14.

Note: The first "y" from left to right in the problem expression text string is a cap y, Y, not lower case y. Mathematica views the two as different characters. The cap Y was changed to lower case prior to solving the problem.

Answer 32

Not quite sure what you mean

Answer 33

hah nevermind i got it you can keep going :)

Answer 34

Cool, so then just like the previous problem the next step is to multiply the to and bottom by the LCD, which in this case is ab

Answer 35

*top

Answer 36

so (ab-ab)/(0)?

Answer 37

how do you figure out the denominator?

Answer 38

I got (a-b)ab for the numerator and b-a for the denominator

Answer 39

To get the denominator: (1/a-1/b)ab (the LCD)=ab/a-ab/b=b-a

Answer 40

ok got it so how did you get the numerator?

Answer 41

I just multiplied (a-b) by the LCD: ab and got (a-b)ab

Answer 42

oh duh lol so would you simplify that farther?

Answer 43

Yes, at this point the equation looks like [(a-b)ab]/(b-a) and if we factor out -1 from the denominator we get [(a-b)ab]/[-(a-b)]

Answer 44

Then the (a-b) cancels and we get -ab

Answer 45

oh ok thank you soo much!!

Answer 46

No problem, glad I could help! :)

Answer 47

\[\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x y}}=\frac{\frac{-x^2+y^2}{x^2 y^2}}{\frac{x^2+2 x y+y^2}{x^2 y^2}}=\frac{-x^2+y^2}{x^2+2 x y+y^2}=\frac{(-(x-y) )}{(x+y)}=\frac{-x+y}{x+y}\]\[1-\frac{2 x}{x+y}=\frac{-x+y}{x+y} \]