document.write( "Question 1209388: Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 2 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 3 fractions, and their product equals 10. What is the value of the first fraction he wrote? \n" ); document.write( "
Algebra.Com's Answer #848730 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The problem is solved far more easily using trial and error than by using formal algebra. However, it is a good exercise for formal algebra.

\n" ); document.write( "Let x be the numerator of the first fraction. Then

\n" ); document.write( "\"%28x%2F%28x-2%29%29%28%28x%2B1%29%2F%28x-1%29%29%28%28x%2B2%29%2F%28x%29%29=10\"

\n" ); document.write( "\"%28x%5E3%2B3x%5E2%2B2x%29%2F%28x%5E3-3x%5E2%2B2x%29=10\"
\n" ); document.write( "\"x%5E3%2B3x%5E2%2B2x=10x%5E3-30x%5E2%2B20x\"
\n" ); document.write( "\"9x%5E3-33x%5E2%2B18x=0\"
\n" ); document.write( "\"3x%283x%5E2-11x%2B6%29=0\"
\n" ); document.write( "\"3x%283x-2%29%28x-3%29=0\"

\n" ); document.write( "\"x=0\" or \"x=2%2F3\" or \"x=3\"

\n" ); document.write( "x=0 is not valid because it results in a division by 0; x=2/3 satisfies the equation but is not an integer; the solution is therefore x=3. The first fraction is then x/(x-2) = 3/1.

\n" ); document.write( "ANSWER: 3/1

\n" ); document.write( "CHECK: (3/1)(4/2)(5/3) = 60/6 = 10

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