document.write( "Question 54617: 3/2x-2 + 1/2 = 2/x-1\r
\n" ); document.write( "\n" ); document.write( "Note the restrictions.\r
\n" ); document.write( "\n" ); document.write( "Thanks! \n" ); document.write( "
Algebra.Com's Answer #36861 by tutorcecilia(2152)\"\" \"About 
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3/2x-2 + 1/2 = 2/x-1
\n" ); document.write( "Since the denominator cannot equal to zero, set each denominator equal to zero and solve for the restriction:
\n" ); document.write( "3/2x-2
\n" ); document.write( "(2x-2)=0
\n" ); document.write( "2x-2=0
\n" ); document.write( "2x=2
\n" ); document.write( "x=1
\n" ); document.write( "So, when x=1, the denominator equals zero which is not acceptable:
\n" ); document.write( "3/2x-2
\n" ); document.write( "3/2(1)-2
\n" ); document.write( "3/2-2
\n" ); document.write( "3/0 [Unacceptable because you cannot divide by zero]
\n" ); document.write( "So the first restriction is that x cannot equal 1
\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "Try the other fraction containing x in the denominator:
\n" ); document.write( "2/x-1
\n" ); document.write( "(x-1)=0
\n" ); document.write( "x-1=0
\n" ); document.write( "x=1
\n" ); document.write( "So, when x=1, the denominator equals zero which also is not acceptable:
\n" ); document.write( "2/x-1
\n" ); document.write( "2/(1)-1
\n" ); document.write( "2/0 [Cannot divide by zero, so, in this fraction also, x cannot equal to one]
\n" ); document.write( ".
\n" ); document.write( "The restrictions are any real number, except that x cannot equal (1).
\n" ); document.write( " \n" ); document.write( "
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