thats pretty easy,i saw it on first try :D... just multiply the top and bottom of the first fraction by the denomiator of the second, and multiply the top and bottom of teh second fraction by the donimator of the first. now you can fuse the two fractions together since they have a common denomiator.. an this common denomiator is the same as teh denominator of the fraction on the right side of the equation.. saw you can cross it out.. (divide both sides of the equation by the same denominator).. your left with a pretty siumple equation...

and then you end up with b(b+2x)=a(a-2x)

you cant simply it more then that

edit: answer = x= (a-b)/2.. if you plug that in in my equation you end up with 0 =0 which is correct XD

first you divide it all with 0, then if you have any negative numbers you naturally put them to a square root and tada, the result is 1=-2, which is a totally suitable answer.

oh my look at all those supersmart maths superheros in here "OH ROFL FUCKING EASY LOL NO TIME TO SOLVE IT BUT ROFL HOW CAN U NOT UNDERSTAND ROFLFOFLFLFLFflflflFLL"

start by finding the LCM ect ect

i know the right solution already...

fucking holidays...i forgot everything :( :D

don't know anymore..

shitmaths

->

((a+x)² + (b-x)²) / (ab - ax + bx - x²) = ...

->

find x in: ab - ax + bx - x² = 0

if there exist any, these x are not allowed.

find x in: (a+x)² + (b-x)² = 2a² - 2b²

This is the answerthats pretty easy,i saw it on first try :D... just multiply the top and bottom of the first fraction by the denomiator of the second, and multiply the top and bottom of teh second fraction by the donimator of the first. now you can fuse the two fractions together since they have a common denomiator.. an this common denomiator is the same as teh denominator of the fraction on the right side of the equation.. saw you can cross it out.. (divide both sides of the equation by the same denominator).. your left with a pretty siumple equation...

and then you end up with

b(b+2x)=a(a-2x)you cant simply it more then that

edit: answer = x= (a-b)/2.. if you plug that in in my equation you end up with 0 =0 which is correct XD

b(b+2x)=a(a-2x)

<=> b² + 2bx = a² - 2ax

<=> 2ax + 2bx = a² - b²

<=> 2x(a+b) = (a+b)(a-b)

<=> 2x = (a-b)

<=> x = (a-b)/2

x= +/- inf

∨

x=(a-b)/2

∨

a+b=0

(a+x)^2-(b-x)^2=2(a^2-b^2)

=>

x=(a-b)/2

ps. btw i know ur math teacher...