document.write( "Question 439897: In a certain fraction, the denominator is 4 less than the numerator. If 3 is added to both the numerator and the denominator, the resulting fraction is equivalent to 3/3. What was the original fraction? Please help! \n" ); document.write( "
Algebra.Com's Answer #304000 by MathLover1(20849)\"\" \"About 
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the denominator \"y\" is 4 less than the numerator \"x\". \r
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\n" ); document.write( "\n" ); document.write( "\"x%2Fy=x%2F%28x%2B4%29\"\r
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\n" ); document.write( "\n" ); document.write( "If 3 is added to both the numerator and the denominator,the resulting fraction is equivalent to 3/3. \r
\n" ); document.write( "\n" ); document.write( "\"%28x%2B3%29%2F%28x%2B4%2B3%29=3%2F3\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x%2B3%29%2F%28x%2B7%29=3%2F3\"..cross multiply\r
\n" ); document.write( "\n" ); document.write( "\"3%28x%2B3%29=3%28x%2B7%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"3x%2B9=3x%2B21%29\"\r
\n" ); document.write( "\n" ); document.write( "\"3x-3x=21-9%29\"...........\"x\" cancels; so, there is no solution
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