# Trick to change subject of formula ?

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Jan 6, 2018

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A couple of ideas...

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There is no one method that will work in all circumstances, but here are a couple of ideas...

Simplifying rational functions

Suppose you are asked to make $x$ the subject of:

$y = \frac{3 x + 2}{x - 5}$

The nuisance here is that $x$ occurs twice. If you leave the right hand expression in that form then you could proceed as follows:

Multiply both sides by $\left(x - 5\right)$ and multiply out to get:

$x y - 5 y = 3 x + 2$

Gather the terms in $x$ on one side by adding $5 y - 3 x$ to both sides to get:

$x y - 3 x = 5 y + 2$

That is:

$x \left(y - 3\right) = 5 y + 2$

Then divide both sides by $y - 3$ to get:

$x = \frac{5 y + 2}{y - 3}$

Alternatively, we can simplify the rational expression first as follows:

$y = \frac{3 x + 2}{x - 5} = \frac{3 \left(x - 5\right) + 17}{x - 5} = 3 + \frac{17}{x - 5}$

Then unwind $x$ from its position to find:

$x = \frac{17}{y - 3} + 5$

Solving polynomials

This can get somewhat messy.

For example, suppose we are given:

$y = {x}^{3} + x$

This is a strictly monotonically increasing function with a well defined inverse, but finding that inverse (i.e. making $x$ the subject) is messy...

Let $x = u + v$

Then our equation becomes:

$y = {\left(u + v\right)}^{3} - \left(u + v\right) = {u}^{3} + {v}^{3} + \left(3 u v - 1\right) \left(u + v\right)$

We can eliminate the term in $\left(u + v\right)$ by adding the constraint:

$3 u v - 1 = 0$

Hence:

${u}^{3} - y + \frac{1}{27 {u}^{3}} = 0$

So:

$27 \left({u}^{3}\right) - 27 y \left({u}^{3}\right) + 1 = 0$

Then by the quadratic formula:

${u}^{3} = \frac{27 y \pm \sqrt{{\left(- 27 y\right)}^{2} - 4 \left(27\right) \left(1\right)}}{2 \cdot 27}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{9 y \pm \sqrt{3 {y}^{2} + 12}}{18}$

Note that this is real valued and our method was symmetrical in $u$ and $v$. Hence we can deduce:

$x = \sqrt[3]{\frac{9 y + \sqrt{3 {y}^{2} + 12}}{18}} + \sqrt[3]{\frac{9 y - \sqrt{3 {y}^{2} + 12}}{18}}$

Ouch!

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