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.
 A: First do the easy simplification on the lefthand side:
$$25\left(\frac{\sqrt{10}-2\sqrt5}{50}\right)=\frac{25}{50}\left(\sqrt{10}-2\sqrt5\right)=\frac12\left(\sqrt{10}-2\sqrt5\right)\;,$$
so the equation can be rearranged to
$$5b=\sqrt5-\frac12\left(\sqrt{10}-2\sqrt5\right)\;.$$
Now divide through by $5$ to get
$$b=\frac{\sqrt5}5-\frac1{10}\left(\sqrt{10}-2\sqrt5\right)=\frac{\sqrt5}5-\frac{\sqrt{10}-2\sqrt5}{10}\;.$$
A: Divide by $5$, then subtract the term $$5\left(\frac{\sqrt{10} - 2 \sqrt 5}{50}\right)$$ from the both sides of the equation:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
$$\iff 5 \left(\frac{\sqrt{10} - 2 \sqrt{5}}{50}\right) + b = \frac {\sqrt 5}{5}\tag{divide by 5}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf -} 5 \left(\frac{\color{blue}{\bf \sqrt{10}  - 2 \sqrt{5}}}{50}\right)\tag{subtract term to left of b}$$
$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf +} \color{red}{\bf 5} \left( \frac{\color{blue}{\bf 2 \sqrt{5} - \sqrt{10}}}{\color{red}{\bf 50}}\right) \tag{"reversal of sign"}$$
$$\iff b = \frac{\sqrt 5}{5} + \left(\frac{2 \sqrt 5 - \sqrt{10}}{10}\right)\tag{cancel common factor 5}$$
A: $$
\begin{align}
25\left(\dfrac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b &= \sqrt{5} \\
5b &= \sqrt{5} - \left(\dfrac{\sqrt{10}-2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} - \left(\dfrac{- ( - \sqrt{10} + 2\sqrt{5} )}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ - \sqrt{10} + 2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \\
b &= \dfrac{1}{5} \cdot \left( \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \right) \\
b &= \dfrac{\sqrt{5}}{5} + \dfrac{2\sqrt{5} - \sqrt{10}}{10}
\end{align}
$$
A: The distributive law is that a(b+c)=ab+ac. For example, 5(3+2)=5*3+5*2.
You can use that for division too, by treating a/b as a*(1/b).
