Simplifying $\frac{1/(\frac{1}{z_1}(1-t)+\frac{1}{z_2}t) - z_1}{(z_2 - z_1)}$ This drives me mad! I am not very good in math but thought I could at least do basic things like this one, but can't figure it out and I spent a day on it. I am trying to simplify:
$\dfrac{\dfrac{1}{\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t} - z_1}{(z_2 - z_1)}$
In Wolfram Alpha it shows me that the solution is:
$\dfrac{tz_1}{t(z_1-z_2)+z_2}$
Which I know is right, but I simply can't figure out the steps to get there. If someone could please help me. I have been as far as doing:
$\dfrac{\dfrac{1 - z_1(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)} {\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t}}{(z_2 - z_1)}$
$\dfrac{1 - z_1(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)} {(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)(z_2 - z_1)}$
And then develop the terms from there, etc. but I just can't get to the equation:
$\dfrac{tz_1}{t(z_1-z_2)+z_2}$
 A: $$\begin{align}
\dfrac{\dfrac{1}{\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t} - z_1}{(z_2 - z_1)} & = \dfrac{\dfrac{z_1z_2}{z_2(1-t)+z_1t} - z_1}{(z_2 - z_1)}\\
& = \dfrac{\dfrac{z_1z_2 - z_1(z_2(1-t) + z_1t)}{z_2(1-t) + z_1t}}{z_2 - z_1}\\
& = \dfrac{\dfrac{z_1z_2t - z_1^2t}{z_2(1-t) + z_1t}}{z_2 - z_1}\\
& = \dfrac{\dfrac{z_1t(z_2 - z_1)}{z_2(1-t) + z_1t}}{z_2 - z_1}\\
& = \dfrac{z_1t}{z_2(1-t) + z_1t}\\
& = \dfrac{z_1t}{t(z_1 - z_2) + z_2}\\
\end{align}$$
A: Starting from where you left off, multiply the top and bottom by $z_1z_2$ to get rid of the fractions, then expand the numerator:
$$
\require{cancel}
\begin{align}
\dfrac{1 - z_1(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)} {(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)(z_2 - z_1)} &= \dfrac{z_1z_2\Big(1 - z_1(\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)\Big)} {z_1z_2\Big((\dfrac{1}{z_1}(1-t)+\dfrac{1}{z_2}t)(z_2 - z_1)\Big)} \\
&= \dfrac{z_1z_2 - z_1(z_2(1-t)+z_1t)}{(z_2(1-t)+z_1t)(z_2 - z_1)} \\
&= \dfrac{\cancel{z_1z_2} - \cancel{z_1z_2} + z_1z_2t - z_1^2t}{(z_2(1-t)+z_1t)(z_2 - z_1)} \\
&= \dfrac{z_1t(z_2 - z_1)}{(z_2(1-t)+z_1t)(z_2 - z_1)}
\end{align}$$
Now cancel the common factor and expand the denominator.
