document.write( "Question 665756: If both the numerator and the denominator of a certain fraction are increased by 1, it becomes 5/6.if both the numerator and denominator are each decreased by 1 it becomes 4/5.find the original fraction \n" ); document.write( "
Algebra.Com's Answer #414116 by ankor@dixie-net.com(22740)\"\" \"About 
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If both the numerator and the denominator of a certain fraction are increased by
\n" ); document.write( " 1, it becomes 5/6.if both the numerator and denominator are each decreased by 1
\n" ); document.write( " it becomes 4/5.find the original fraction
\n" ); document.write( ":
\n" ); document.write( "\"If both the numerator and the denominator of a certain fraction are increased by 1, it becomes 5/6.\"
\n" ); document.write( "\"%28%28n%2B1%29%29%2F%28%28d%2B1%29%29\" = \"5%2F6\"
\n" ); document.write( "Cross multiply
\n" ); document.write( "6(n+1) = 5(d+1)
\n" ); document.write( "6n + 6 = 5d + 5
\n" ); document.write( "6n - 5d = 5 - 6
\n" ); document.write( "6n - 5d = -1
\n" ); document.write( ":
\n" ); document.write( "\" both the numerator and denominator are each decreased by 1 it becomes 4/5.\"
\n" ); document.write( "\"%28%28n-1%29%29%2F%28%28d-1%29%29\" = \"4%2F5\"
\n" ); document.write( "Cross multiply
\n" ); document.write( "5(n-1) = 4(d-1)
\n" ); document.write( "5n - 5 = 4d - 4
\n" ); document.write( "5n - 4d = -4 + 5
\n" ); document.write( "5n - 4d = 1
\n" ); document.write( ":
\n" ); document.write( "We can use elimination here, Mult the 1st eq by 4, mult the 2nd eq by 5
\n" ); document.write( "Subtract the 1st equation from the 2nd equation
\n" ); document.write( "25n - 20d = 5
\n" ); document.write( "24n - 20d = -4
\n" ); document.write( "--------------- Subtraction eliminates d, find n
\n" ); document.write( "n = 9
\n" ); document.write( ":
\n" ); document.write( "find d using 5n - 4d = 1
\n" ); document.write( "5(9) - 4d = 1
\n" ); document.write( "45 - 4d = 1
\n" ); document.write( "-4d = 1 - 45
\n" ); document.write( "-4d = 44
\n" ); document.write( "d = +11
\n" ); document.write( ":
\n" ); document.write( "\"9%2F11\" is the original equation, you check this for yourself
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