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ik its choice a and b but ineed the restriction?
\[\frac{x^2+4x-45}{x^2+10x+9}=\frac{x-5}{x+1}\]

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The first step is to factor the numerator and denominator.
is the restriction -9
\[\mathrm{Factor}\:x^2+4x-45:\quad \left(x+9\right)\left(x-5\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{x^2+10x+9}\] \[\mathrm{Factor}\:x^2+10x+9:\quad \left(x+9\right)\left(x+1\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[\mathrm{Cancel\:the\:common\:factors:}\:x+9\] \[=\frac{x-5}{x+1}\] that is the answer
@jammy987 are you understand
|dw:1439493910772:dw|
you do not want factor to be 0 so restrictions are the values that will make the factor = 0
restrictions x not = -1, not = 5, not = -9
so option A is correct do not divide by 0
You don't want the denominator to be zero. The numerator can be zero. These two factors cannot be zero: \(x + 9 \ne 0\) \(x + 1 \ne 0\) x - 5 .can be zero because a zero in the numerator is permitted.

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