document.write( "Question 77159: what value for x must be excluded in the following fraction?\r
\n" ); document.write( "\n" ); document.write( "x-3
\n" ); document.write( "over
\n" ); document.write( "(4x-5)(x+1)
\n" ); document.write( "am i right to say
\n" ); document.write( "5/4,1,3
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Algebra.Com's Answer #55451 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Given:
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\n" ); document.write( "\"%28x-3%29%2F%28%284x-5%29%28x%2B1%29%29\"
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\n" ); document.write( "What values of x are not allowed?
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\n" ); document.write( "The only thing that is not allowed in this expression is division by zero. Therefore,
\n" ); document.write( "neither of the two terms in the denominator can equal zero. You can solve for values
\n" ); document.write( "of x that are not allowed by setting each of the factors in the denominator equal to zero
\n" ); document.write( "and then solving for the corresponding value of x.
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\n" ); document.write( "So begin by saying:
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\n" ); document.write( "\"4x+-+5+=+0\"
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\n" ); document.write( "Solve this by adding 5 to both sides to get \"4x+=+5\" and then dividing by the multiplier
\n" ); document.write( "of x ... 4 ... to get that \"x+=+5%2F4\". So you cannot let x equal \"5%2F4\" because
\n" ); document.write( "if it did, then the factor \"4x-5\" would equal zero. That one you got correctly.
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\n" ); document.write( "Next do the same type of calculation for the other factor in the denominator \"x%2B1\".
\n" ); document.write( "Set it equal to zero ...
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\n" ); document.write( "\"x%2B1+=+0\" and then subtract +1 from both sides to get \"x+=+-1\". This means that
\n" ); document.write( "if you let x = -1, the factor \"x+%2B+1\" becomes zero and that would create a division
\n" ); document.write( "by zero which is not an allowable situation in algebra. This one you called +1 ...
\n" ); document.write( "suggesting that you probably made a sign error somewhere.
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\n" ); document.write( "As far as the numerator going to zero ... that is not a problem. So you can solve the
\n" ); document.write( "\"x+-3+=+0\" and find that if \"x+=+%2B3\" the numerator goes to zero, but the factors in
\n" ); document.write( "the denominator are not zero when x = +3. There is no division by zero that occurs when
\n" ); document.write( "x = 3 ... and when x = 3 the two factors in the denominator are:
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\n" ); document.write( " and since the numerator
\n" ); document.write( "is zero when x = 3, the expression becomes:
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\n" ); document.write( "\"0%2F28\" and this is just zero ... which is OK.
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\n" ); document.write( "In summary, the two non-allowed values are \"x+=+5%2F4\" and \"x+=+-1\"
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\n" ); document.write( "Hope this helps you to understand the problem a little more.
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