Question

show working to solve for x:
x/m+n=x/n+m
and get x=-mn

Answer 1

\[{x \over m}+n={x \over n}+m\]Is this the problem

Answer 2

yes

Answer 3

\[{x \over m}+n={x \over n}+m\]\[{x \over m}-{x \over n}=m-n\]\[{xn \over mn}-{xm \over mn}=m-n\]\[{xn-xm \over mn}=m-n\]\[{x(n-m) \over mn}=m-n\]\[x={(m-n)mn \over (n-m)}\]Getting close :)

Answer 4

so far im with you...

Answer 5

\[x={(m-n)mn \over (n-m)}\]\[-1(x)=-1({(m-n)mn \over (n-m)})\]\[-x={(n-m)mn \over (n-m)}\]\[-x=mn\]\[x=-mn\]

Answer 6

any questions?

Answer 7

if i was given the question, how would i know to continue like that?

Answer 8

if i wasn't given the answer

Answer 9

So basically what was my mind set? Ok..

Answer 10

I knew I eventually needed x by itself. So my first thing I wanted to do was get the x's on the same side of the equation

Answer 11

I moved the constants to the other side as well

Answer 12

In order to subtract those fractions and bring the x's together I needed a common denominator

Answer 13

if u were told just to solve for x, without knowing its going to be -mn, would you have still gotten -mn?

Answer 14

Honestly I probably would have left it at the end of my first post

Answer 15

ok, thanks.
i really appreciate your time.

Answer 16

I wonder if \[\frac{ m-n }{ n-m }=-1\]for all cases. If you knew that then I guess you can arrive at that final answer

Answer 17

right!

Answer 18

Also you could realize that if you multiplied it by -1 you could cancel it out. the denominator. I don't really know why you would continue on without knowing these other things.