Solving for $b$ in $25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$ What are the steps to get from:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
to:
$$b = \frac{\sqrt{5}}{5} + \frac{2\sqrt{5} - \sqrt{10}}{10}$$
Thanks.

$$
\begin{align}
25\left(\dfrac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b &= \sqrt{5} \\
5b &= \sqrt{5} - \left(\dfrac{\sqrt{10}-2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} - \left(\dfrac{- ( - \sqrt{10} + 2\sqrt{5} )}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ - \sqrt{10} + 2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \\
b &= \dfrac{1}{5} \cdot \left( \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \right) \\
b &= \dfrac{\sqrt{5}}{5} + \dfrac{2\sqrt{5} - \sqrt{10}}{10}
\end{align}
$$
