document.write( "Question 1023155: In a certain positive fraction the denominator is greater than the numerator by 3 if 1 is subtracted from both the numerator and denominator the fraction is decreased by 1/14.find the fraction \n" ); document.write( "
Algebra.Com's Answer #638748 by fractalier(6550)![]() ![]() You can put this solution on YOUR website! Call the original fraction \n" ); document.write( " \n" ); document.write( "Then we add one to top and bottom and get \n" ); document.write( " \n" ); document.write( "This fraction is 1/14 less than the original...thus we have to solve \n" ); document.write( " \n" ); document.write( "Now multiply everything by the lowest common denominator... \n" ); document.write( " \n" ); document.write( "and we get \n" ); document.write( "14x(x+2) - (x+2)(x+3) = 14(x+3)(x-1) \n" ); document.write( "14x^2 + 28x - x^2 - 5x - 6 = 14x^2 + 28x - 42 \n" ); document.write( "The 14x^2 and 28x cancel out... \n" ); document.write( "-x^2 - 5x - 6 = -42 \n" ); document.write( "x^2 + 5x + 6 = 42 \n" ); document.write( "x^2 + 5x - 36 = 0 \n" ); document.write( "(x + 9)(x - 4) = 0 \n" ); document.write( "x = -9 and x = 4 \n" ); document.write( "The original fraction must be \n" ); document.write( "4/7. \n" ); document.write( " |