Question

simplify each:
please explain how to do it rather than just giving me the answer

Answer 1

ok what is it :)

Answer 2

\[(1/y^2 - 1/x^2) / (x/y) +2+(y/x)\]

Answer 3

@callie2240 Got this?

Answer 4

Or whoever; I got to go.

Answer 5

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

Answer 6

@phi seems to have it :)

Answer 7

or is it
\[ \frac{ \frac{1}{y^2} - \frac{1}{x^2}}{\frac{x}{y} + 2 + \frac{y}{x} } \]

Answer 8

the 2nd one

Answer 9

if the 2nd way, you should put parens around the bottom so we know:

(1/y2−1/x2)/ ( (x/y)+2+(y/x) )

Answer 10

oh I meant to sorry

Answer 11

The first step is get rid of the complicated bottom

Answer 12

and to do that we should simplify it to a single fraction

so for the time being, let's concentrate on
\[ \frac{x}{y} + 2 + \frac{y }{x} \]

any idea what the common denominator should be ?

Answer 13

xy

Answer 14

yes, Can you add up these terms, using xy as the common denominator ?

Answer 15

yes hold on my computer is being stupid

Answer 16

like this:
\[ \frac{x}{y} \cdot \frac{x}{x} + 2 \cdot \frac{xy}{xy} + \frac {y}{x} \cdot \frac{y}{y}\]

Answer 17

\[x^2/xy+2xy/xy+y^2/xy\]

Answer 18

yes, which we can write as
\[ \frac{x^2 + 2xy + y^2}{xy} \]

we add the tops, and put that over the common denominator.

now we use this idea:
\[ \frac{a}{b} \cdot \frac{b}{a}= \frac{\cancel{a}}{\cancel{b}} \cdot \frac{\cancel{b}}{\cancel{a}}= 1\]

Answer 19

in other words, if we multiply the complicated fraction's bottom by
\[ \frac{xy}{x^2 + 2xy + y^2} \]

it will turn into 1, and we can ignore it.
of course, that means we have to multiply the top by the same thing

Answer 20

ok lets do it

Answer 21

got it

Answer 22

\[ \frac{ \frac{1}{y^2} - \frac{1}{x^2}}{\frac{x}{y} + 2 + \frac{y}{x} } = \frac{ \frac{1}{y^2} - \frac{1}{x^2}}{\frac{x^2+2xy+y^2}{xy}}\]
now multiply top and bottom
\[ \frac{ \frac{1}{y^2} - \frac{1}{x^2}\cdot \frac{xy}{x^2+2xy+y^2}}{\frac{x^2+2xy+y^2}{xy}\cdot \frac{xy}{x^2+2xy+y^2}} \]

Answer 23

\[xy/(x^2y+2xy^2+y^4) - xy/(x^4+2x^2y+xy^2)\] all over 1

Answer 24

there should be parens in that mess
\[ \frac{ (\frac{1}{y^2} - \frac{1}{x^2})\cdot \frac{xy}{x^2+2xy+y^2}}{\frac{x^2+2xy+y^2}{xy}\cdot \frac{xy}{x^2+2xy+y^2}} \]
the bottom is 1
\[ \frac{ (\frac{1}{y^2} - \frac{1}{x^2})\cdot \frac{xy}{x^2+2xy+y^2}}{1} \\= (\frac{1}{y^2} - \frac{1}{x^2})\cdot \frac{xy}{x^2+2xy+y^2}\]

Answer 25

yes, but I would not multiply things. If there is any hope of simplifying it, we should factor it, and hope somethings will cancel.

so instead, keep it like this
\[ (\frac{1}{y^2} - \frac{1}{x^2})\cdot \frac{xy}{x^2+2xy+y^2} \]

now subtract the two terms 1/y^2 - 1/x^2

Answer 26

common denominator is x^2 y^2

Answer 27

ok

Answer 28

\[x^2/x^2y^2 - y^2/x^2y^2\]

Answer 29

oh sorry forgot parenthesis

Answer 30

hold on

Answer 31

yes, or (x^2 - y^2)/ (x^2 y^2)

Answer 32

can i multiply now?

Answer 33

you do want to put the top over a common denominator
\[ \frac{(x^2 -y^2)} {x^2 y^2} \]

Answer 34

so far, we have
\[ \frac{(x^2 -y^2)} {x^2 y^2} \cdot \frac{xy}{x^2+2xy+y^2} \]

multiplying will give you the correct, (but unsimplified) answer.
when you see a mess, think "factor"

can you factor x^2 - y^2 (difference of squares has a factors nicely)

Answer 35

(x-y)(x+y)

Answer 36

can you factor the bottom x^2 + 2xy + y^2

hint: for a made up problem (unlike the real world) you might expect one of the factors up top to be in the bottom, so that things will cancel

Answer 37

(x+y)(x+y)

Answer 38

so we have
\[ \frac{(x-y)(x+y)x y }{ x^2 y^2 (x+y)(x+y) }\]
cancel time !

Answer 39

wait what just happened? did we multiply?

Answer 40

yes, top times top and bottom times bottom... not too tricky

Answer 41

ok

Answer 42

so is this just the xy fraction on the left? or is it the whole thing?

Answer 43

we got to
\[ \frac{(x^2 -y^2)} {x^2 y^2} \cdot \frac{xy}{x^2+2xy+y^2} \]
then you factored to get
\[ \frac{(x -y)(x+y)} {x^2 y^2} \cdot \frac{xy}{(x +y)(x+y)} \]

then (I should have let you do this step, because you like to multiply so much) :
\[ \frac{ (x -y)(x+y) xy } { x^2y^2(x +y)(x+y)} \]

Answer 44

ok and yeah it's fun

Answer 45

now cancel terms that are in the top and bottom (that divide out to become 1)

Answer 46

[(x-y)xy]/[x^2y^2(x+y)]

Answer 47

did you forget how to simplify x/x^2 = x/(x*x) ?

Answer 48

no I just forgot to do it

Answer 49

x-y/ (xy)(x+y)

Answer 50

yes though (x-y)/ (xy(x+y) ) is better x -y*stuff is different from (x-y)*stuff
because the 2nd way means stuff multiplies both the x and the -y

Answer 51

or divides in this case

Answer 52

so is that it? am I done?

Answer 53

yes

Answer 54

yay! thank you! do you think you could help me with one more?

Answer 55

if you go faster.

Answer 56

i'll try..sorry