document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=533726>Question 533726</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The denominator of a fraction is an integer that is one greater than the numerator. A second fraction is the reciprocal of the integer that is one less than the numerator of the first fraction. If the product of two fractions is 3/8, identify the first fraction. </SPAN>\n" );
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document.write( "Read the words, and write exactly what they say.\r<BR>\n" );
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document.write( "The denominator of a fraction is an integer that is one greater than the numerator.  \r<BR>\n" );
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document.write( "We don't know what the numerator is, so let's call it x.  The denominator is one greater than the numerator, so it's x+1.  Thus this fraction is x/(x+1).  Both x and x+1 must be integers.  Otherwise the denominator won't be an integer.\r<BR>\n" );
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document.write( "A second fraction is the reciprocal of the integer that is one less than the numerator of the first fraction.  \r<BR>\n" );
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document.write( "The numerator of the first fraction is x.  An integer one less than x is x-1.  The reciprocal is 1 over the thing.  So this fraction is 1/(x+1).\r<BR>\n" );
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document.write( "If the product of two fractions is 3/8, identify the first fraction.\r<BR>\n" );
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document.write( "Product means multiply.\r<BR>\n" );
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document.write( "x/(x+1) * 1/(x-1) = 3/8\r<BR>\n" );
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document.write( "Multiply numerator by numerator and denominator by denominator.\r<BR>\n" );
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document.write( "x*1 / (x+1) * (x-1) = 3/8\r<BR>\n" );
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document.write( "The numerator is x.  The denominator is a binomial multiplication.  So use FOIL.\r<BR>\n" );
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document.write( "(x-1) * (x+1)   = \r<BR>\n" );
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document.write( "F - first times first:  x*x=x^2<BR>\n" );
document.write( "O - outside times outside:  1*x = x<BR>\n" );
document.write( "I - inside times inside:  -1*x=-x<BR>\n" );
document.write( "L - last times last:  -1*1=-1\r<BR>\n" );
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document.write( "Add those terms:  x^2 + x - x - 1 = x^2 - 1\r<BR>\n" );
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document.write( "Thus the fraction is x/(x^2-1)=3/8\r<BR>\n" );
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document.write( "Cross multiple whenever one fraction equals another.  8 * x = 3 (x^2-1)\r<BR>\n" );
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document.write( "Multiply 3 by both terms in the ()\r<BR>\n" );
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document.write( "8x = 3x^2 - 3  \r<BR>\n" );
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document.write( "Bring everything to one side of the equation.\r<BR>\n" );
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document.write( "0 = 3x^2 - 8x - 3  Remember any equation can be turned around because we can multiply both sides by -1 and then add each side to both sides.\r<BR>\n" );
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document.write( "-1*0 = -1 * (3x^2 - 8x - 3) = -3x^2 + 8x + 3  \r<BR>\n" );
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document.write( "Add 3x^2 to both sides, subtract 8x from both sides, subtract 3 from both sides.\r<BR>\n" );
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document.write( "0 + 3x^2 -8x - 3 = -3x^2 + 3x^2 -8x +8x -3 + 3\r<BR>\n" );
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document.write( "3x^2 - 8x - 3 = 0\r<BR>\n" );
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document.write( "Reverse foil to get two terms, and then set each equal to 0.\r<BR>\n" );
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document.write( "(3x   )(x   )  we need terms that when one is multiplied by 3 and then they are added, are -8, and when multiplied together = -3.  Factors of -3 are 3 and -1, -3 and 1.  Try 1 and -3.\r<BR>\n" );
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document.write( "(3x+1)(x-3)=0  FOIL proves that works.  Set each = to 0 and solve for x.\r<BR>\n" );
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document.write( "3x+1=0  subtract 1 from both sides.  3x+1-1=0-1 --->  3x = -1\r<BR>\n" );
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document.write( "Divide by 3 --->  x=-1/3   That won't work, because x has to be an integer.\r<BR>\n" );
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document.write( "x-3 = 0  add 3 to both sides.  x-3+3=0+3 --->  x=3.\r<BR>\n" );
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document.write( "X=3 is the answer.  Check by substituting it into the original equation we built.\r<BR>\n" );
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document.write( "3/4 * 1/2 = 3/8\r<BR>\n" );
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document.write( "3*1 = 3 (the numerator) and 4*2=8 (the denominator).\r<BR>\n" );
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document.write( "3/8=3/8  HURRAH!!!!!!!!!\r<BR>\n" );
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document.write( " </SPAN>\n" );
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