Question

Did I do this right?

Answer 1

\[\frac{ x-\frac{ x }{ y }}{ y-\frac{ y }{ x }} =\frac{ \frac{ xy }{ y } -\frac{ x }{y} }{\frac{ yx }{x }-\frac{ y }{ x }}=\frac{ \frac{ xy-x }{ y } }{ \frac{ yx-x }{ y } }=\frac{ xy-x }{ y }*\frac{ x }{ yx-y }=\frac{ -xx }{-yy}\]

Answer 2

i think you got a typo

Answer 3

where?

Answer 4

\[=\frac{ \frac{ xy-x }{ y } }{ \frac{ yx-x }{ y } }\]

Answer 5

yep.... that bottom y is an x... the rest I transcriber right though

Answer 6

the denominator of the denominator should be \(x\)
then there is a mistake on the next liine

Answer 7

Yeah nice catch sat \[\frac{ a }{ b } \pm \frac{ c }{ d } = \frac{ ad \pm bc }{ bd }\]

Answer 8

your life would be easier if, instead of doing all the work top and bottom, multiply all by \(xy\) top and bottom as a first step

Answer 9

it looks like you cancelled something at the last line, but nothing cancels there

Answer 10

\[\frac{ xy-x }{ y }*\frac{ x }{ yx-y }\neq\frac{ -xx }{-yy}\]

Answer 11

\[\frac{ x-\frac{ x }{ y }}{ y-\frac{ y }{ x }}\times \frac{xy}{xy}\] might be an easier way to go

Answer 12

oh shoot.. so \[\frac{ xy-x ^{2} }{ y ^{2}x-m }\] for the answer? I did cancel everything out at the end

Answer 13

close

Answer 14

ok... i didn't learn that way... how did you choose xy to multiply both by?

Answer 15

get rid of the pesky extra denominators

Answer 16

your way works as well, just don't cancel anything

Answer 17

i see.. since x is the extra denominator on top multiply both sides by x and since y is the extra denominator on the bottom multiply both sides by y also?

Answer 18

\[\frac{ x^2y-x ^{2} }{ y ^{2}x-y^2}\]
yeah right

Answer 19

ok let me work it out again to see if i get what you did. 1 sec...

Answer 20

kk

Answer 21

can you show me what you get after multiplying the original equation by xy/xy... I keep messing that part up I think

Answer 22

Never mind I figured it out . Thanks!