ParthKohli
If\[x = \dfrac{4ab}{a + b}\]Then prove that\[\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} = 2\]
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ParthKohli
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@UnkleRhaukus
ParthKohli
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I can't believe I'm finding 9th grade so hard. :'-|
ParthKohli
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Ahahahahaha well! I got it.
.Sam.
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lol
UnkleRhaukus
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how?
ParthKohli
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Do you want to know my proof?
UnkleRhaukus
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is it elegant?
ParthKohli
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\[x = \dfrac{4ab}{a+b} \qquad \Rightarrow \qquad \dfrac{x}{2a}=\dfrac{2ab}{a+b} ~~\text{and} ~~ \dfrac{x}{2b} = \dfrac{2a}{a+b}\]I don't know, it may be elegant.
UnkleRhaukus
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i see what your doing there...
ParthKohli
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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
ParthKohli
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Oh, and in my earlier post, I meant to write\[\dfrac{x}{2a} = \dfrac{2b}{a+b}\]
ParthKohli
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Simplifying, we get\[\dfrac{x + 2a}{x - 2a} = \dfrac{3b+a}{b-a}\]and\[\dfrac{x + 2b}{x-2b} =\dfrac{3a+b}{a - b}\]
ParthKohli
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So now we can add both equations
ParthKohli
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\[\begin{aligned}\dfrac{x + 2a}{x - 2a} + \dfrac{x+2b}{x-2b} &=\dfrac{3b+a}{b-a} + \dfrac{3a+b}{a-b} \\ \\ \\ & = \dfrac{-3b-a+3a+b}{a-b} \\ \\ \\ &= \dfrac{2a-2b}{a-b} \\ \\ \\ & = 2\end{aligned} \]
ParthKohli
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Phew. QED
UnkleRhaukus
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this is 9th grade?
ParthKohli
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Yeah.
ParthKohli
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No wait, this is 8th grade. I was just reviewing stuff.