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.

Divide by $5$, then subtract the term $$5\left(\frac{\sqrt{10} - 2 \sqrt 5}{50}\right)$$ from the both sides of the equation:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
$$\iff 5 \left(\frac{\sqrt{10} - 2 \sqrt{5}}{50}\right) + b = \frac {\sqrt 5}{5}\tag{divide by 5}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf -} 5 \left(\frac{\color{blue}{\bf \sqrt{10}  - 2 \sqrt{5}}}{50}\right)\tag{subtract term to left of b}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf +} \color{red}{\bf 5} \left( \frac{\color{blue}{\bf 2 \sqrt{5} - \sqrt{10}}}{\color{red}{\bf 50}}\right) \tag{"reversal of sign"}$$
$$\iff b = \frac{\sqrt 5}{5} + \left(\frac{2 \sqrt 5 - \sqrt{10}}{10}\right)\tag{cancel common factor 5}$$
