document.write( "Question 599990: The numerator of a fraction is 2 more than the denominator. If denominator is doubled at the numerator is decreased by 1, the sum of the original fraction and the new one is 7/3. Find the original fraction. \n" ); document.write( "
Algebra.Com's Answer #379148 by ankor@dixie-net.com(22740)\"\" \"About 
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The numerator of a fraction is 2 more than the denominator.
\n" ); document.write( " If denominator is doubled and the numerator is decreased by 1, the sum of the original fraction and the new one is 7/3.
\n" ); document.write( " Find the original fraction.
\n" ); document.write( ":
\n" ); document.write( "Let x = the denominator
\n" ); document.write( "then
\n" ); document.write( "(x+2) = the numerator
\n" ); document.write( ":
\n" ); document.write( "\"%28%28x%2B2%29%29%2Fx\" = the original fraction
\n" ); document.write( ":
\n" ); document.write( "the equation for the statement:
\n" ); document.write( "\"If denominator is doubled and the numerator is decreased by 1, the sum of the original fraction and the new one is 7/3.\"
\n" ); document.write( "\"%28%28x%2B1%29%29%2F%282x%29\" + \"%28%28x%2B2%29%29%2Fx\" = \"7%2F3\"
\n" ); document.write( "Multiply by 6x
\n" ); document.write( "6x*\"%28%28x%2B1%29%29%2F%282x%29\" + 6x*\"%28%28x%2B2%29%29%2Fx\" = 6x*\"7%2F3\"
\n" ); document.write( "cancel the denominators and you have
\n" ); document.write( "3(x+1) + 6(x+2) = 2x(7)
\n" ); document.write( "3x + 3 + 6x + 12 = 14x
\n" ); document.write( "9x + 15 = 14x
\n" ); document.write( "15 = 14x - 9x
\n" ); document.write( "15 = 5x
\n" ); document.write( "x = 15/5
\n" ); document.write( "x = 3
\n" ); document.write( ":
\n" ); document.write( "\"5%2F3\" is the original fraction
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