Question

Please help and i'll medal =)

If 3xm+2ym-2yn-3xn=0 and "m" does not equal "n," then what is the value of y in terms of x?

Answer 1

\[3xm+2ym-2yn-3xn=0\] and your job it so solve for \(y\) right?

Answer 2

yes

Answer 3

ok leave everything with a y on the left, everything else on the right
\[2ym-2yn=3xn-3xm\]

Answer 4

factor out the y on the left
\[y(2m-2n)=3xn-3xm\]

Answer 5

ohhhh okay

Answer 6

then divide to finish with
\[y=\frac{3xn-3xm}{2m-2n}\]

Answer 7

what about those 2 x's

Answer 8

and m cannot equal n

Answer 9

i hope i got the m and n straight

Answer 10

the m and the n has to disappear

Answer 11

the reason they wrote that m cannot equal n is because of your answer
if m was equal to n the denominator would be zero and you cannot divide by zero
they just wrote it to make sure the answer makes sense

Answer 12

oh ic

Answer 13

ok we can make them go away i think, hold on

Answer 14

\[y=\frac{3xn-3xm}{2m-2n}\] factor again as
\[y=\frac{3x(n-m)}{3(m-n)}\]

Answer 15

typo there should be a 2 in the denominator

Answer 16

bottom is 2 but yea so do they cancel out that way?

Answer 17

\[y=\frac{3x(n-m)}{2(m-n)}\]

Answer 18

shouldnt this be a negative

Answer 19

they do cancel, but they don't leave a 1 they cancel to get -1 because
\[\frac{n-m}{m-n}=-1\]

Answer 20

final answer if i did not mess up with the m and n is
\[y=-\frac{3}{2}x\]

Answer 21

hmmm kinda confusing

Answer 22

yeah i agree

Answer 23

tryna figure out why the m and n's went to the left side

Answer 24

oh because you wanted so solve for \(y\) so all the terms with y have to be on one side of the equal sign, everything else on the other

Answer 25

hmmmm okay i got it im gonna have to go over this a couple more times but thanks satellite

Answer 26

yw
we can do it again slowly if you like, starting with \[3xm+2ym-2yn-3xn=0\]

Answer 27

yea thatd be great

Answer 28

ok it is clear you want an answer that looks like \(y=something\) right?

Answer 29

yea

Answer 30

\[3xm+2ym-2yn-3xn=0\] the middle two terms have a y in them , the first and last terms do not

Answer 31

yea so faactor it out right

Answer 32

we leave the middle two terms on the left, and subtract \(3xm\) from both sides, also add \(3xn\) to both sides

Answer 33

that is how i got to the line
\[2ym-2yn=3xn-3xm\]
ok to here?

Answer 34

yea

Answer 35

yep

Answer 36

now factor out the y on the left, so you can see what do divide both sides by
\[y(2m-2n)=3xn-3xm\]

Answer 37

now divide both sides by \(2m-2n\) and get \[y=\frac{3xn-3xm}{2m-2n}\]

Answer 38

hey can i try something else

Answer 39

sure

Answer 40

what about m(3x+2y)-n(2y+3x=0

Answer 41

but i dont understand why it becomes just (3x+2y) (m-n)=0 where does that other (3x+2y) go?

Answer 42

that is right, but i am not sure how it is going to get you to \(y=expression\)

Answer 43

you really need to have all the y on one side, everything else on the other

Answer 44

yea

Answer 45

this is really confusing i am screwed if i get this on the GRE

Answer 46

once we get to
\[y=\frac{3xn-3xm}{2m-2n}\] we have achieved that goal
now it is a matter of factoring and cancelling on the right

Answer 47

factor a \(3x\) from the numerator and a \(2\) from the denominator, get
\[y=\frac{3x(n-m)}{2(m-n)}\]

Answer 48

ok im with you

Answer 49

then cancel, get
\[y=-\frac{3x}{2}\]

Answer 50

the gimmick here is to recognize that
\[\frac{n-m}{m-n}=-1\]

Answer 51

so you treat the (n-m) as a variable and you divide from each other and cancel out right?

Answer 52

yeah, it is always the case that
\[\frac{a-b}{b-a}=-1\] since the denominator is the negative of the numerator and vice versa

Answer 53

ohhh..... is that rule?

Answer 54

it is just a fact

Answer 55

i did not know this....

Answer 56

\[\frac{5-2}{2-5}=\frac{3}{-3}=-1\] for example
you can try it with your own numbers, see that it works

Answer 57

hmmmmm okay i think i got it now...

Answer 58

ohhhhhh.... right omfg that blew my mind

Answer 59

thank so much satellite

Answer 60

yw, good luck with the gre

Answer 61

thanks!