Hi,
According to wolfram alpha x+y / x-y can also be represented as [-(2x)/(y-x)] -1
When it is in this form it works for me I just can't seem to get it in that form,
any help (or suggestions or basic rules to apply) would be greatly appreciated.
Hi,
According to wolfram alpha x+y / x-y can also be represented as [-(2x)/(y-x)] -1
When it is in this form it works for me I just can't seem to get it in that form,
any help (or suggestions or basic rules to apply) would be greatly appreciated.
We want to express the express (x+y)/(x-y) as [ -2x / (y-x) ] - 1.
In order to form it into that form we will first take out -1 common from the denominator to get
(x+y)/(x-y) = - ( x+y)/ (y-x) . Now divide the numerator by the denominator, quotient being 1 and remainder 2x.
Thus we get: (x+y)/(x-y) = - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] = -1 - 2x / ( y-x)
Thank you for the quick reply, but I have no idea how you get from:
"Now divide the numerator by the denominator, quotient being 1 and remainder 2x.
Thus we get: - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] = -1 - 2x / ( y-x)"
I haven't done algebra in 4-5 years, could you please explain this step by step? (as I have forgotten "basic math rules")
It's - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] that I don't understand.
Hello, CHUBBS!
According to Wolfram Alpha: can also be written
Their answer is correct . . . but really stupid!
I tried to figure out their reasoning . . . weird!
We have: .
Multiply by
Long division:
. .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Why not divide "normally"? .
. .
Therefore: .