# How to find postfix of this equation 5+2/[3*{6+2+(5-3/2)-5}+2]-1*5 step by step?

Apr 6, 2018

5 enter 2 enter 3 enter 6 enter 2 + 5 enter 3 enter 2 / - + 5 - * 2 + / + 1 enter 5 * -

#### Explanation:

$5 + \frac{2}{3 \cdot \left\{6 + 2 + \left(5 - \frac{3}{2}\right) - 5\right\} + 2} - 1 \cdot 5$

The question asks for Postfix. I know this as Reverse Polish Notation which places entered results in a memory STACK. Provided that this memory stack has room for at least 6 elements, you can write this expression as

5 enter 2 enter 3 enter 6 enter 2 + 5 enter 3 enter 2 / - + 5 - * 2 + / + 1 enter 5 * -

But be careful! If you use a standard RPN calculator for this with only 4 stack elements, you will get the wrong answer!

This should simplify to $\frac{4}{43}$.

$5 + \frac{2}{3 \cdot \left\{6 + 2 + \left(\textcolor{red}{\frac{10}{2}} - \frac{3}{2}\right) - 5\right\} + 2} - 1 \cdot 5$

$5 + \frac{2}{3 \cdot \left\{6 + 2 + \textcolor{red}{\frac{7}{2}} - 5\right\} + 2} - 1 \cdot 5$

$5 + \frac{2}{3 \cdot \left\{\textcolor{red}{8} + \frac{7}{2} - 5\right\} + 2} - 1 \cdot 5$

$5 + \frac{2}{3 \cdot \left\{\textcolor{red}{3} + \frac{7}{2}\right\} + 2} - 1 \cdot 5$

$5 + \frac{2}{3 \cdot \left\{\textcolor{red}{\frac{6}{2}} + \frac{7}{2}\right\} + 2} - 1 \cdot 5$

$5 + \frac{2}{3 \cdot \textcolor{red}{\frac{13}{2}} + 2} - 1 \cdot 5$

$5 + \frac{2}{\textcolor{red}{\frac{39}{2}} + 2} - 1 \cdot 5$

$5 + \frac{2}{\frac{39}{2} + \textcolor{red}{\frac{4}{2}}} - 1 \cdot 5$

$5 + \frac{2}{\textcolor{red}{\frac{43}{2}}} - 1 \cdot 5$

$5 + \textcolor{red}{\frac{4}{43}} - 1 \cdot 5$

$5 + \frac{4}{43} - \textcolor{red}{5}$

$\frac{4}{43}$