"Solving" an equation through division by a variable

classic Classic list List threaded Threaded
2 messages Options
Reply | Threaded
Open this post in threaded view
|

"Solving" an equation through division by a variable

MarkAU
Hello Carl,

In Precalculus, you make clear a pitfall of dividing both sides of an equation by a variable if there is a possibility that that variable might equal zero – for instance:

x^2 = x
==> x = 1 (dividing both sides by x)

which leads to a valid solution for x, but if we do it this way:

x^2 = x
x^2 - x = 0
x(x - 1) = 0
==> x = 1 or x = 0

... thus revealing a previously "hidden" second solution of zero for x.

For some of the exercises in Precalculus for certain topics, I've taken the trouble to document for myself the workings to their solutions where I'm concerned that I might later forget how to get started solving that particular kind of problem.

Among others, I've done that for problems 57-96 on page 931 of Precalculus v3 (converting equations between rectangular and polar coordinates).

In problems 61-63, for the sake of brevity, I took the shortcut of dividing both sides of a resulting polar equation by r on the understanding that a value of r=0 was extraneous because it would not lead us to a DEFINED value for angle theta as asked for, eg:

y = -x      (problem 61)
r sin (theta) = -r cos (theta)
sin (theta) = -cos (theta)
tan (theta) = -1
==> theta = 3pi / 4

In problems 68-76, I similarly took the shortcut of dividing both sides of a resulting polar equation by r on the understanding that a solution of zero for r was unwanted, eg:

x^2 + y^2 = x      (problem 72)
r^2 = r cos (theta)
==> r = cos (theta)

I did put a note at the top of my document pointing out that those understandings were to be taken into account.

Putting aside for the moment your role as an examiner in insisting on full methodological exactitude being demonstrated in your students' exam papers, are there any other lurking pitfalls in dividing both sides of an equation by a variable PROVIDED THAT a value of zero cannot occur for that variable; or that a value of zero for that variable would be extraneous or unwanted; AND that you understand the preconditions and implications of taking such a shortcut?
Reply | Threaded
Open this post in threaded view
|

Re: "Solving" an equation through division by a variable

Carl Stitz
Administrator
Thanks for the note!

If I understand your question correctly - you are correct.   You are free to divide both sides by a variable when solving an equation as long as you either take that variable being equal to zero in a separate case or argue that for the purposes of your calculations, the variable is never zero or the zero case is taken into account in a more general solution.

I appreciate your careful reading!

Carl