Question 77159
Given:
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{{{(x-3)/((4x-5)(x+1))}}}
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What values of x are not allowed?
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The only thing that is not allowed in this expression is division by zero.  Therefore,
neither of the two terms in the denominator can equal zero. You can solve for values
of x that are not allowed by setting each of the factors in the denominator equal to zero
and then solving for the corresponding value of x.
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So begin by saying:
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{{{4x - 5 = 0}}}
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Solve this by adding 5 to both sides to get {{{4x = 5}}} and then dividing by the multiplier
of x ... 4 ... to get that {{{x = 5/4}}}.  So you cannot let x equal {{{5/4}}} because
if it did, then the factor {{{4x-5}}} would equal zero. That one you got correctly.
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Next do the same type of calculation for the other factor in the denominator {{{x+1}}}.
Set it equal to zero ...
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{{{x+1 = 0}}} and then subtract +1 from both sides to get {{{x = -1}}}. This means that
if you let x = -1, the factor {{{x + 1}}} becomes zero and that would create a division
by zero which is not an allowable situation in algebra. This one you called +1 ... 
suggesting that you probably made a sign error somewhere.
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As far as the numerator going to zero ... that is not a problem. So you can solve the
{{{x -3 = 0}}} and find that if {{{x = +3}}} the numerator goes to zero, but the factors in
the denominator are not zero when x = +3. There is no division by zero that occurs when
x = 3 ... and when x = 3 the two factors in the denominator are:
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{{{(4x-5)*(x+1) = (4*3 - 5)*(3+1) = (12 - 5)*(4) = 7*4 = 28}}} and since the numerator
is zero when x = 3, the expression becomes:
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{{{0/28}}} and this is just zero ... which is OK.
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In summary, the two non-allowed values are {{{x = 5/4}}} and {{{x = -1}}}
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Hope this helps you to understand the problem a little more.