# 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.

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Divide by $5$, then subtract the term $$5\left(\frac{\sqrt{10} - 2 \sqrt 5}{50}\right)$$ from the both sides of the equation:

$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$

$$\iff 5 \left(\frac{\sqrt{10} - 2 \sqrt{5}}{50}\right) + b = \frac {\sqrt 5}{5}\tag{divide by 5}$$

$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf -} 5 \left(\frac{\color{blue}{\bf \sqrt{10} - 2 \sqrt{5}}}{50}\right)\tag{subtract term to left of b}$$

$$\iff b = \frac{\sqrt 5}{5} \color{blue}{\bf +} \color{red}{\bf 5} \left( \frac{\color{blue}{\bf 2 \sqrt{5} - \sqrt{10}}}{\color{red}{\bf 50}}\right) \tag{"reversal of sign"}$$

$$\iff b = \frac{\sqrt 5}{5} + \left(\frac{2 \sqrt 5 - \sqrt{10}}{10}\right)\tag{cancel common factor 5}$$

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Does this make sense, Cypras? – amWhy Mar 4 '13 at 23:40
How can you just reverse the sign? – Cypras Mar 4 '13 at 23:44
Because we can multiply the last term through by -1: $\color{blue}{\bf -} 5 \left(\frac{\color{blue}{\bf \sqrt{10} - 2 \sqrt{5}}}{50}\right) = +5 \cdot -1 \left(\frac{\color{blue}{\bf \sqrt{10} - 2 \sqrt{5}}}{50}\right) = 5 \left(\frac{\color{blue}{\bf 2 \sqrt{5} -\sqrt{10} }}{50}\right)$ – amWhy Mar 4 '13 at 23:48
$-(\sqrt {10} - 2\sqrt 5) = -\sqrt{10} + 2\sqrt 5 = (2\sqrt 5 - \sqrt{10})$ – amWhy Mar 4 '13 at 23:52
can't say I understand? Why do you swap $2\sqrt(10)-2\sqrt(5)$ around? – Cypras Mar 4 '13 at 23:52

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

$$b=\frac{\sqrt5}5-\frac1{10}\left(\sqrt{10}-2\sqrt5\right)=\frac{\sqrt5}5-\frac{\sqrt{10}-2\sqrt5}{10}\;.$$

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can you explain your first simplification please? – Cypras Mar 4 '13 at 23:30
@Cypras: It’s a general fact of the arithmetic of fractions that $$a\cdot\frac{b}c=\frac{a}1\cdot\frac{b}c=\frac{ab}c=\frac{a}c\cdot\frac{b}1= \frac{a}c\cdot b\;;$$ now take $a=25$, $b=\sqrt{10}-2\sqrt5$, and $c=50$. – Brian M. Scott Mar 4 '13 at 23:33
Why do you not have to divide $\sqrt(10)-2\sqrt(5)$ by 5 when dividing through by 5? – Cypras Mar 4 '13 at 23:38
@Cypras: I have to divide the term $\frac12\left(\sqrt{10}-2\sqrt5\right)$ by $5$, and I did, when I changed $\frac12$ to $\frac1{10}$. Dividing by $5$ is the same as multiplying by $\frac15$, and $\frac15\cdot\frac12\cdot x=\frac1{10}\cdot x$, no matter what $x$ is. – Brian M. Scott Mar 4 '13 at 23:41

\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}

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\sqrt{10} to get $\sqrt{10}$. – Git Gud Mar 4 '13 at 23:31

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).

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