SOLUTION: state the excluded values of the variables for each rational expression. it would be really helpfully if u can answer my question plzzzzzz....i really need help.thank you soooo muc

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: state the excluded values of the variables for each rational expression. it would be really helpfully if u can answer my question plzzzzzz....i really need help.thank you soooo muc      Log On


   



Question 123085: state the excluded values of the variables for each rational expression. it would be really helpfully if u can answer my question plzzzzzz....i really need help.thank you soooo much..
22 n to the 2nd and then underlined with 11n and if u can solve a few more questions of this kind and longer i would be really glad.
thank you again

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
As I interpret your problem, you were given:
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22n%5E2%2F%2811n%29
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and you are to find the excluded variables.
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In this problem you could make the mistake of simplifying the problem by dividing the 11 in the
denominator into the 22 of the numerator and then dividing the n in the denominator into
the n-squared of the numerator to reduce the expression to:
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22n%5E2%2F%2811n%29+=+%2822%2F11%29%2A%28n%5E2%2Fn%29+=+2%2An
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Then it would look as if n could be any real value. But that is not the case. The real problem
you have here is with the n in the denominator. If the n in the denominator is zero, then
you have a division by zero ... and in algebra a division by zero is not allowed.
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If you took n%5E2%2Fn and allowed n to be zero, the problem would become zero divided by zero
and that is not allowed. If you did the division first, you would have n%5E2%2Fn+=+n and then
if you let n = 0 it would appear OK. But it is not OK because simplifying it first hid the
fact that a division by zero occurred.
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So, summing it all up ... n cannot be zero in this problem. Other than that, n can be any
real number.
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Hope this helps you to understand the problem.
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