zero on top in rational equations - help explain please

[Jand]

Here is the equation (had to find the value of x):

2x + 3 - 2x - 8 = 1
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x - 4 ww 2x + 1

I did the usual stuff and ended up with

Well (31x-25)/((x-4)(2x+1))=0

I got stuck and used an on-line solver to find value(s) of x - it says X₁ = 25/31

If you put x=25/31 into the simplified equation (where I got stuck) the top = zero, and the bottom = -8.3444.....

Why is the top a valid solution for x but the bottom (with x in it) ignored?

is the top being = to 0 a special case?

Does x with the subscript 1 mean something special?

I can't find anything about it on the net - so I'm asking for help here.


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[stapel]

Are you allowed to divide by zero? So is a "solution" which causes division by zero a valid solution?

When is a fraction equal to zero? Under this condition, does it matter what is the value of the denominator (other than it not being zero itself also)?

 

[Jand]

Are you allowed to divide by zero? So is a "solution" which causes division by zero a valid solution?

When is a fraction equal to zero? Under this condition, does it matter what is the value of the denominator (other than it not being zero itself also)?

Thanks - you set me on the right track to find out.

I think the solution relies on defining inequalities from a rational function. Here is a good video that explains it well:

https://www.youtube.com/watch?v=UXuSv-4WlAg


Then I found this that is even simpler:

http://www.mathsisfun.com/algebra/rational-expression.html


"To find the roots of a Rational Expression we only need to find the the roots of the top polynomial, so long as the Rational Expression is in 'Lowest Terms'."

[FONT=Verdana, Arial, Tahoma, sans-serif]I'm going to have to look at it all a bit longer to really understand what is going on - but as a simple guide I can remember that when the denominator is zero the value of the variable is undefined and never a root (and that the variable in the numerator and denominator live in different sets on the number line because they are constituent parts of different polynomials).[/FONT]