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: $\displaystyle \frac{x+y}{x-y}$ can also be written $\displaystyle \frac{-2x}{y-x}-1$
Their answer is correct . . . but really stupid!
I tried to figure out their reasoning . . . weird!
We have: .$\displaystyle \frac{x+y}{x-y}$
Multiply by $\displaystyle \tfrac{-1}{-1}\!:\;\;\frac{-1}{-1}\cdot\frac{x+y}{x-y} \;=\;\frac{-x-y}{-x+y} \;=\;\frac{-y-x}{y-x}$
Long division:
. . $\displaystyle \begin{array}{ccccc}&&&-1 \\ && -- & -- \\ y-x & | & -y & -x \\ && -y & +x \\ && -- & -- \\ &&& -2x\end{array}$
Therefore: .$\displaystyle \frac{-y-x}{y-x} \;=\;-1 - \frac{2x}{y-x}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Why not divide "normally"? .$\displaystyle \frac{x+y}{x-y}$
. . $\displaystyle \begin{array}{cccccc} &&&&1 \\ && --&--&-- \\ x-y & | & x & + & y \\ && x & -& y \\ && --&--&-- \\ &&&& 2y \end{array}$
Therefore: .$\displaystyle \frac{x+y}{x-y} \;=\;1 + \frac{2y}{x-y}$