Problem approximating for a chemical problem: Solve using suitable approximation by hand: 
$$s(s-\alpha)=5\times10^{-13} \\ \frac{\alpha}{(s-\alpha)(0.4-2\alpha)^2}=10^8$$
Now to solve this I tried assuming $s\approx\alpha\approx0.2$ and then tried finding $s$ but failed any help?
 A: The product of those is:
$$\frac{s\alpha}{(0.4-2\alpha)^2}=5\times10^{-5}$$
The first term is almost zero so, $s\approx0,s-\alpha\approx0\implies s\approx\alpha$.
So:
$$5\times10^{-5}=\frac{s\alpha}{(0.4-2\alpha)^2}\approx\frac{s^2}{0.16}$$
This makes $s\approx\sqrt{5\times10^{-5}\times0.16}\approx2.828\times10^{-3}$.

The actual values(considering quadratic/cubic) are $$s\approx2.78899\times10^{-3},\alpha\approx2.78898\times10^{-3}$$ not so far away.
A: What you did is very correct. If I may comment, you could have done a little more even by hand; considering, as you properly did, that $(0.4-2\alpha) \approx 0.4$, the two equations write $$s (s-\alpha )=5 \times 10^{-13}$$ $$ \frac{\alpha}{s-\alpha}=16 \times 10^6$$ and, as you wrote, their product gives $\alpha s=8 \times 10^{-6}$. So, reusing the first equation, $$s^2-\alpha s=5 \times 10^{-13}$$ then $$s^2=5 \times 10^{-13}+8 \times 10^{-6}=8 \times 10^{-6}\times\Big(1+\frac{5 \times 10^{-13}}{8 \times 10^{-6}}\Big)$$ and using, for small $x$, $\sqrt{1+x} \approx 1+\frac x 2$, then $$s \approx \sqrt{8 \times 10^{-6}}\times\Big(1+\frac{5 \times 10^{-13}}{2 \times 8 \times 10^{-6}}\Big)\approx 0.002828427213$$ and from $\alpha s=8 \times 10^{-6}$, then $$\alpha \approx 0.002828427036$$ while the exact solutions of the original problem would have been $s=0.002788985013$ and $\alpha=0.002788984833$; from this last result $s-\alpha\approx 1.79 \times 10^{-10}$ while the previous series of approximations leads to $s-\alpha\approx 1.77 \times 10^{-10}$.
By the way, eliminating $\alpha$ from the first equation, thoe original problem transforms to a quartic polynomial in $s$.
