is $\frac{a-b}{a+b} - \frac{c-d}{c+d}$ equal to $ \frac{a-c}{a+c}$?

Question

I am using values out of a cross-correlation analysis and my intuition tells me that the equation in the title is true, however I would like to prove it. $$ \frac{a-b}{a+b} - \frac{c-d}{c+d} = \frac{a-c}{a+c}$$

Unfortunately my math skills are horrible and I could not find this equation on-line (which now makes me think that my intuition is faulty).

Could you at least point out what mental strategy you would follow to try to solve it?

Thanks for any help

PS= any suggestion to improve my way of asking questions is also welcome.

all the best

Answer 1

$$\frac{a-b}{a+b}-\frac{c-d}{c+d}=$$ $$\frac{(a-b)(c+d)}{(a+b)(c+d)}-\frac{(c-d)(a+b)}{(c+d)(a+b)}=$$ $$\frac{(a-b)(c+d)-(c-d)(a+b)}{(a+b)(c+d)}=$$ $$\frac{2(ad-bc)}{(a+b)(c+d)}$$

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If you would like to look if your identity is correct, set your variables:

Choose $a=2,b=3,c=4,d=5$:

$$ \frac{2-3}{2+3} - \frac{4-5}{4+5} = \frac{2-4}{2+4}$$ $$ \frac{-1}{5} - \frac{-1}{9} = \frac{-2}{6}$$ $$ -\frac{1}{5}+\frac{1}{9} = -\frac{1}{3}$$ $$ -\frac{4}{45} = -\frac{1}{3}$$

So you know that:

$$ -\frac{4}{45} \ne -\frac{1}{3}$$

So your identity is not correct

Answer 2

First, we can assume that: $$ (a+b)\neq 0\qquad c+d\neq 0\qquad a+c\neq 0 $$ Rewrite the left-hand side: $$ \frac{(a-b)(c+d)-(a+b)(c-d)}{(a+b)(c+d)}=\frac{ac+ad-bc-bd-ac+ad-bc+bd}{(a+b)(c+d)}= $$ $$ =2\frac{ad-bc}{(a+b)(c+d)} $$ So we need the following to hold: $$ 2\frac{ad-bc}{(a+b)(c+d)}=\frac{a-c}{a+c} $$ Which is generally not true, and leads to the equation: $$ a^2 (c-d)+a \left(3 b c+b d-c^2-3 c d\right)+b c^2-b c d=0 $$ If this is true for eg. all $a$, then all the coefficients should be $0$: $$ c-d=3 b c+b d-c^2-3 c d=bc(c-d)=0 $$ From $c=d$ we have $3 b c+b d-c^2-3 c d=4 (b - d) d$, so either $d=0$ or $b=d$.