Solving a seemingly simple equation

Question

$$\frac{1}{\mu_1}[Ah+h^2]=\frac{1}{\mu_2}[-Bh+h^2]$$

We have that $A=B$. Without any explination the next line is: $$A=B=\frac{(\mu_1 - \mu_2)}{(\mu_1 + \mu_2)}h$$ I feel like there is some sort of trick and I don't know how with my standard rearranging technique. Can someone teach me how please?

Answer 1

Use the fact $A=B$ and rewrite

$$\frac{1}{\mu_1}(A+h) = \frac{1}{\mu_2}(-A+h),$$ or $$\mu_2 A + \mu_2 h = -\mu_1 A + \mu_1 h.$$ Then, $$(\mu_1 + \mu_2)A = (\mu_1- \mu_2)h,$$ or, $$A = \frac{\mu_1 - \mu_2}{\mu_1 + \mu_2}h.$$ Note, I used $h\neq 0$ to derive the first equation.

Answer 2

We have $\frac{1}{\mu_1}[Ah+h^2]=\frac{1}{\mu_2}[-Bh+h^2]$

$A=B$

$ \frac{[Bh+h^2]}{\mu_1}=\frac{[-Bh+h^2]}{\mu_2}$

$\frac{B}{\mu_1}+\frac h\mu_1 = -\frac B\mu_2+\frac h\mu_2 $

$B\bigg[\frac1\mu_1+\frac1\mu_2\bigg] = h\bigg[\frac1\mu_2-\frac1\mu_1\bigg]$

$B \bigg[\frac{\mu_1+\mu_2}{\mu_1\mu_2}\bigg]=h\bigg[\frac{\mu_1-\mu_2}{\mu_1\mu_2}\bigg]$

$B =\bigg[\frac{\mu_1-\mu_2}{\mu_1+\mu_2}\bigg]h$

$A=B =\bigg[\frac{\mu_1-\mu_2}{\mu_1+\mu_2}\bigg]h$

Answer 3

HINT: for $$A=B$$ and $$h\neq 0$$ we get the equation $$\frac{A+h}{\mu_1}=\frac{-A+h}{\mu_2}$$ Now use the hint above!