Algebra 101: Making subject I am trying to figure out how the upper equation can be equivalent to the one at the bottom. 



My best guess is that something is being multiplied or divided from both sides but cannot figure out what. 

I proceed as follows:

I first expand everything and cancel out similar terms to ensure that I have L+W on one side and I get the following:

$$ rL(\rho-1)&gt;= -{\sigma}{\tau}(L+W)$$

So isolating L+W gets a very different result. 

I'd really appreciate if anyone can guide me in the right direction.
Continuing from where you left off, plus assuming $1 \gt \rho \implies 1 - \rho \gt 0$, $L + W \gt 0$ and $1-\frac{\sigma\tau}{r(1-\rho)} \gt 0$, gives

$$\begin{equation}\begin{aligned}
rL(\rho-1) &amp; \ge -\sigma\tau(L+W) \\
-\frac{rL(1 - \rho)}{L+W} &amp; \ge -\sigma\tau \\
-\frac{L}{L+W} &amp; \ge -\frac{\sigma\tau}{r(1-\rho)} \\
1-\frac{L}{L+W} &amp; \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
\frac{W}{L+W} &amp; \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
W &amp; \ge (L + W)\left(1-\frac{\sigma\tau}{r(1-\rho)}\right) \\
\frac{W}{1-\frac{\sigma\tau}{r(1-\rho)}} &amp; \ge L+W \\
L + W &amp; \le \frac{W}{1-(\sigma\tau)/(r(1-\rho))}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Continuing from where you left off, plus assuming $1 \gt \rho \implies 1 - \rho \gt 0$, $L + W \gt 0$ and $1-\frac{\sigma\tau}{r(1-\rho)} \gt 0$, gives

$$\begin{equation}\begin{aligned}
rL(\rho-1) &amp; \ge -\sigma\tau(L+W) \\
-\frac{rL(1 - \rho)}{L+W} &amp; \ge -\sigma\tau \\
-\frac{L}{L+W} &amp; \ge -\frac{\sigma\tau}{r(1-\rho)} \\
1-\frac{L}{L+W} &amp; \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
\frac{W}{L+W} &amp; \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
W &amp; \ge (L + W)\left(1-\frac{\sigma\tau}{r(1-\rho)}\right) \\
\frac{W}{1-\frac{\sigma\tau}{r(1-\rho)}} &amp; \ge L+W \\
L + W &amp; \le \frac{W}{1-(\sigma\tau)/(r(1-\rho))}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$