Tips for solving equations for a variable

Question

I am trying to solve the following equation for $V_{12}$:

$$\sqrt{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}\sqrt{\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}}=\sqrt{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}$$

(this is not really relevant, but it is the derivation of the formula for the relativistic addition of velocities using the Doppler k-factor as presented in this paper).

For me as a highschool student, this isn't an easy equation to solve. Of course, I would square everything to remove the square roots. Then I thought that it might be good to have $V_{12}$ (the variable I'm trying to solve for) on one side and the two fractions with $V_1$ and $V_2$ on the oder side. This way, I got

$$\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}=\frac{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}$$

However, I am a bit clueless what to do next. I thought of multiplying by $\left(1-\frac{V_{12}}{c}\right)$ to get rid of the fraction on the right side, but then I would again have $V_{12}$ on the right side. Whatever I tried seemed not to be helpful (for example, I had the idea to multiply out everything after squaring and then multiply by the denominators to remove the fractions, but this would only lead to overly complicated expressions with even more fractions like $\frac{V_1V_2V_{12}}{c^3}$.

The correct result should be

$$V_{12}=\frac{V_2-V_1}{1-\frac{V_1V_2}{c^2}}$$

but I cannot figure out how to get there.

So: What are gerneral reccomendations and tips to approach solving an equation like this for one variable? (Note that I'm not specifically asking for a derivation of the last mentioned formula from the first since this would probably be outside the scope of this site)

(I also tried using Wolfram Alpha to solve for $V_{12}$ and then analyse the step-by-step solution, but it didn't work)

Answer 1

Ultimately, it looks like your goal is to solve for $x$ in the equation

$$y = \frac{1+x}{1-x}$$

While an oversimplification, I think it should be clear how it applies to your situation. Take $y$ to be your complex fraction, and $x$ to be $V_{12}/c$.

(So I suppose this would be my main tip: boil down your equation into its "essence", in a sense, like the above equation does. Solving for $x$ in the above equation makes it trivial to find $V_{12}$ in your case. A mass of symbols can be daunting and mislead the mind, but by attributing names and definitions to larger structures, you can simplify something messy into something familiar. For instance, this is a quotient of linear functions, whose inverses are well known.)

To find $x$, simply work as follows:

\begin{align*} y = \frac{1+x}{1-x} &\iff y(1-x) = 1+x \\ &\iff y - yx = 1 +x \\ &\iff -x - yx = 1 - y \\ &\iff x(-1-y) = 1 - y \\ &\iff x = \frac{1-y}{-1-y} \\ &\iff x = \frac{y-1}{y+1} \end{align*}

Of course, bear in mind that $x \ne 1$ and $y \ne -1$ throughout this process. (These represent asymptotes of the function and are never attained, and result in divisions by zero when plugged in.)

Answer 2

You're probably making this more complicated than it needs to be. You got this far $$\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}=\frac{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}.$$ Note that by multiplying the all three fractions by $c/c$ we can rewrite it as as $$\frac{c + V_{12}}{c - V_{12}} = \frac{\frac{c + V_{2}}{c - V_{2}}}{\frac{c + V_{1}}{c - V_{1}}}.$$ Rewriting the right-hand side so we don't have two stacked fractions: $$\frac{c + V_{12}}{c - V_{12}} = \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}.$$ Now we multiply the denominator of the left-hand side over and distribute: $$c + V_{12} = c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) - V_{12}\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right).$$ Collect the $V_{12}$ terms on the left-hand side: \begin{align} V_{12} + V_{12}\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) &= c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) - c\\ V_{12}\left(1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) &= c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right), \end{align} which gives us $$V_{12} = \frac{c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right)}{1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}}.$$ At this point it's just a matter of expanding and simplifying the right-hand side to get the result: \begin{align} V_{12} &= \frac{c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right)}{1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}}\\ &=\frac{\frac{(c + V_{2})(c - V_{1}) - (c - V_{2})(c + V_{1})}{(c - V_{2})(c + V_{1})}}{\frac{(c - V_{2})(c + V_{1}) + (c + V_{2})(c - V_{1})}{(c - V_{2})(c + V_{1})}}\\ &=\frac{(c + V_{2})(c - V_{1}) - (c - V_{2})(c + V_{1})}{(c - V_{2})(c + V_{1}) + (c + V_{2})(c - V_{1})}\\ &=\frac{2cV_{2} - 2cV_{1}}{2c^{2} - 2V_{1}V_{2}}\\ &=\frac{V_{2} - V_{1}}{1 - \frac{V_{1}V_{2}}{c^{2}}}. \end{align}