Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

4 Answers 4

up vote 2 down vote accepted

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

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

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

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

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

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

share|improve this answer
    
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

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

share|improve this answer
    
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
1  
@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

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

share|improve this answer
    
\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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.