Do all rational equations have a single solution? Why is that so?

Whenever you solve a rational equation you must make sure the result obtained for
an answer does not allow the denominator of one of the rational expressions to assume a value of ZERO, [ as division by zero is undefined and therefore prohibited.
For example if we have
2x/(x-3) =(x2 -9x)/ x
when we multiply out by x(x-3) we get
2x(x) = (x2 -9x)(x-3)
so 2x2 = x(x-9)(x-3)
2x2 = x(x2 - 12x + 27)
2x2

= x3 - 12x2 + 27x
so
0 = x3 - 14x2 + 27x
0 = x(x2 - 14x + 27)
so solutions are 0 and 7 + v22 and 7 -v22
but 0 makes right hand side expression have zero in denominator so it is not
a solution.
We actually have to look at all obtained solutions to be sure they ae not extraneous.
Suppose we had obtained a 3 for a solution. That would make the left side denominator equal zero and we would have to dismiss that, if 3 was obtained.
The two irrational solutions we have obtained are genuine solutions as neither introduces a zero to a denominator
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Asked 7/10/2012 2:31:47 PM

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