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Dividing by variable when solving algebraic equation
#1
Dividing by variable when solving algebraic equation
I know for all you math masters out there, this is basic stuff. But for people like me, it used to baffle me a little how dividing by a variable could sometimes lead to an incomplete solution set when solving an equation.

Consider the following for example:

x(x+1) = 0

The recommended approach to solving this is to use the zero factor property, leading to the solution set {-1, 0} for x.

But suppose, instead, you divided both sides by x, cancelling out the standalone x's on the left side of the equation. Then you're left with just:

x+1 = 0 which can only lead to the unique solution of -1 for x.
0 is no longer a solution.

I actually searched online for a website that could clearly explained why this happened, but couldn't find one. Maybe I used the wrong search keywords or something.

Anyway, I figured it out yesterday why this happens.

When you divide both sides of the original equation by x, the equation actually changes (meaning: it becomes a different equation with a different solution set for x).

Before dividing the equation by x, 0 was clearly one of the solutions.

But when you divide by x, 0 can no longer be a solution because then that would mean you were dividing by 0 in the case of x = 0! Division by 0 cannot work so x = 0 stops being a solution, and only -1 becomes the acceptable solution.

So rule of thumb when it comes to solving algebraic equations:

Don't divide by the variable when you have the option of using the zero factor property. Or you may end up with an incomplete answer.

I feel so happy finally getting this after all this time.
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Messages In This Thread
Dividing by variable when solving algebraic equation - by Grandizer - October 25, 2016 at 3:35 am

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