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How would you substitute $u=$$\frac{mH}{k-m}$ into $v=(1-u)(u+H)$ to produce $v=\frac{kH}{(k-m)^2}(k-m(1+H))$?

Ideally I'd like a step by step breakdown with the intuition described so I can learn how to approach this kind of question in the future over this question.

K.M.
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1 Answers1

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$1-u= 1- \frac{mH}{k-m}$ & $u+H = \frac{mH}{k-m} +H$. Further simplifying,

$1-u= \frac{k-m(1+H)}{k-m}$ & $u+H = \frac{kH}{k-m}$. Multiply the two expressions, we're done!!

SL_MathGuy
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