Trick to change subject of formula ?

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Jan 6, 2018

Answer:

A couple of ideas...

Explanation:

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 = (3x+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 #(x-5)# and multiply out to get:

#xy-5y = 3x+2#

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

#xy-3x=5y+2#

That is:

#x(y-3) = 5y+2#

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

#x = (5y+2)/(y-3)#

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

#y = (3x+2)/(x-5) = (3(x-5)+17)/(x-5) = 3+17/(x-5)#

Then unwind #x# from its position to find:

#x = 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 = (u+v)^3-(u+v) = u^3+v^3+(3uv-1)(u+v)#

We can eliminate the term in #(u+v)# by adding the constraint:

#3uv-1 = 0#

Hence:

#u^3-y+1/(27u^3) = 0#

So:

#27(u^3)-27y(u^3)+1 = 0#

Then by the quadratic formula:

#u^3 = (27y+-sqrt((-27y)^2-4(27)(1)))/(2 * 27)#

#color(white)(u^3) = (9y+-sqrt(3y^2+12))/18#

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

#x = root(3)((9y+sqrt(3y^2+12))/18)+root(3)((9y-sqrt(3y^2+12))/18)#

Ouch!

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