Question

If\[x = \dfrac{4ab}{a + b}\]Then prove that\[\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} = 2\]

Answer 1

@UnkleRhaukus

Answer 2

I can't believe I'm finding 9th grade so hard. :'-|

Answer 3

Ahahahahaha well! I got it.

Answer 4

lol

Answer 5

how?

Answer 6

Do you want to know my proof?

Answer 7

is it elegant?

Answer 8

\[x = \dfrac{4ab}{a+b} \qquad \Rightarrow \qquad \dfrac{x}{2a}=\dfrac{2ab}{a+b} ~~\text{and} ~~ \dfrac{x}{2b} = \dfrac{2a}{a+b}\]I don't know, it may be elegant.

Answer 9

i see what your doing there...

Answer 10

OK, let me continue.\[\dfrac{x + 2a}{x - 2a} = \dfrac{2b + a + b}{2b - a - b}\]and\[\dfrac{x + 2b}{x-2b} = \dfrac{2a + a + b}{2a - a - b}\]by componendo and dividendo property

Answer 11

Oh, and in my earlier post, I meant to write\[\dfrac{x}{2a} = \dfrac{2b}{a+b}\]

Answer 12

Simplifying, we get\[\dfrac{x + 2a}{x - 2a} = \dfrac{3b+a}{b-a}\]and\[\dfrac{x + 2b}{x-2b} =\dfrac{3a+b}{a - b}\]

Answer 13

So now we can add both equations

Answer 14

\[\begin{aligned}\dfrac{x + 2a}{x - 2a} + \dfrac{x+2b}{x-2b} &=\dfrac{3b+a}{b-a} + \dfrac{3a+b}{a-b} \\ \\ \\ & = \dfrac{-3b-a+3a+b}{a-b} \\ \\ \\ &= \dfrac{2a-2b}{a-b} \\ \\ \\ & = 2\end{aligned} \]

Answer 15

Phew. QED

Answer 16

this is 9th grade?

Answer 17

Yeah.

Answer 18

No wait, this is 8th grade. I was just reviewing stuff.