document.write( "Question 81644This question is from textbook Algebra I
\n" ); document.write( ": How do you solve for x when it is in the denominator on both sides of a rational equation? And what are the restrictions for X? This is the problem: \r
\n" ); document.write( "\n" ); document.write( "4/x = 5/x - 1/2
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Algebra.Com's Answer #58491 by tutorcecilia(2152)\"\" \"About 
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The restrictions in this case is that x cannot equal to zero because you cannot divide by zero.
\n" ); document.write( "\"%284%2Fx%29\" = \"%285%2Fx+-+1%2F2%29\"
\n" ); document.write( "\"%28%282x%294%2Fx%29\" = \"%28%282x%295%2Fx+-+%282x%291%2F2%29\" [find the LCD (2x); multiply each term by the LCD]
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\n" ); document.write( "8=10-x [cancel wherever possible]
\n" ); document.write( "8=10-x [solve for the x-term]
\n" ); document.write( "8-8=10-8-x
\n" ); document.write( "0=2-x
\n" ); document.write( "-2=2-2-x
\n" ); document.write( "-2=-x
\n" ); document.write( "-2/-1=-x/-1
\n" ); document.write( "2=x
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\n" ); document.write( "check by plugging (x=2) back into the original equation and solve:
\n" ); document.write( "\"%284%2Fx%29\" = \"%285%2Fx+-+1%2F2%29\"
\n" ); document.write( "\"%284%2F2%29\" = \"%285%2F2+-+1%2F2%29\"
\n" ); document.write( "2=\"%284%2F2%29\"
\n" ); document.write( "2=2 [checks out]
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