document.write( "Question 100024: in a rational no. twice the numerator is 2 more than the denominator. if 3 is added to each, the numerator or the denominator, the new fraction is 2/3. find the original no. \n" ); document.write( "
Algebra.Com's Answer #72849 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Let x=the numerator
\n" ); document.write( "And let y=the denominator\r
\n" ); document.write( "\n" ); document.write( "Now we are told that:\r
\n" ); document.write( "\n" ); document.write( "2x=y+2-----------------------eq1
\n" ); document.write( "and
\n" ); document.write( "(x+3)/(y+3)=2/3-------------------eq2\r
\n" ); document.write( "\n" ); document.write( "multiply each term in eq2 by 3(y+3) to get rid of fractions and we get:\r
\n" ); document.write( "\n" ); document.write( "3(x+3)=2(y+3) get rid of parens
\n" ); document.write( "3x+9=2y+6 subtract 9 from both sides:\r
\n" ); document.write( "\n" ); document.write( "3x=2y-3---------------------eq2
\n" ); document.write( "multiply eq1 by 2 and subtract it from eq2:
\n" ); document.write( "eq2-2*(eq1)=\r
\n" ); document.write( "\n" ); document.write( " 3x=2y-3
\n" ); document.write( "-4x=-2y-4\r
\n" ); document.write( "\n" ); document.write( "-x=-7 or\r
\n" ); document.write( "\n" ); document.write( "x=7-----------------numerator
\n" ); document.write( "substitute x=7 into eq1\r
\n" ); document.write( "\n" ); document.write( "2*7=y+2 subtract 2 from both sides\r
\n" ); document.write( "\n" ); document.write( "14-2=y\r
\n" ); document.write( "\n" ); document.write( "y=12-----------denominator\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "(7+3)/(12+3)=2/3\r
\n" ); document.write( "\n" ); document.write( "10/15=2/3
\n" ); document.write( "2/3=2/3\r
\n" ); document.write( "\n" ); document.write( "also
\n" ); document.write( "2*7=12+2
\n" ); document.write( "14=14\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor\r
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