document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1111018>Question 1111018</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sum of the reciprocals of two numbers is 7 and the difference of the reciprocals is 3. What are the numbers? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #726012 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=726012\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you have 2 equations that need to be solved simultaneously.\r<BR>\n" );
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document.write( "they are:\r<BR>\n" );
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document.write( "1/A + 1/B = 7\r<BR>\n" );
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document.write( "1/A - 1/B = 3\r<BR>\n" );
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document.write( "add the 2 equations together to get:\r<BR>\n" );
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document.write( "2/A = 10\r<BR>\n" );
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document.write( "solve for A to get A = 2/10 = 1/5\r<BR>\n" );
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document.write( "when A = 1/5, 1/A + 1/B = 7 becomes:\r<BR>\n" );
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document.write( "1/(1/5) + 1/B = 7\r<BR>\n" );
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document.write( "simplify to get 5 + 1/B = 7\r<BR>\n" );
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document.write( "subtract 5 from both sides to get 1/B = 2\r<BR>\n" );
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document.write( "solve for B to get B = 1/2\r<BR>\n" );
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document.write( "you have A = 1/5 and B = 1/2.\r<BR>\n" );
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document.write( "your first equation is 1/A + 1/B = 7\r<BR>\n" );
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document.write( "replace A with 1/5 and B with 1/2 to get:\r<BR>\n" );
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document.write( "1/(1/5) + 1/(1/2) = 7\r<BR>\n" );
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document.write( "simplify to get 5 + 2 = 7, which is true.\r<BR>\n" );
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document.write( "your second equaton is 1/A - 1/B = 3\r<BR>\n" );
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document.write( "replace A with 1/5 and B with 1/2 to get:\r<BR>\n" );
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document.write( "1/(1/5) - 1/(1/2) = 3\r<BR>\n" );
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document.write( "simplify to get 5 - 2 = 3, which is true.\r<BR>\n" );
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document.write( "both equations are true when A = 1/5 and B = 1/2, therefore the solution can be assumed to be good.\r<BR>\n" );
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document.write( "how do you get 5 out of 1/(1/5)?\r<BR>\n" );
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document.write( "it's a simple matter of multiplying both numerator and denominator by 5.\r<BR>\n" );
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document.write( "1/(1/5) * 5/5 = (5*1) / (5*1/5) = 5/1 = 5\r<BR>\n" );
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document.write( "you can do this without changing the fraction because multiplying by 5/5 is the same as multiplying by 1 which keeps the value of the original fraction the same.\r<BR>\n" );
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document.write( "to confirm, used your calculator to divide 1 by (1/5) and you will see that the answer is 5.\r<BR>\n" );
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document.write( "the same procedure to solve these simultaneous equations is used to solve simultaneous equations that do not include fractions.\r<BR>\n" );
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document.write( "since the first equation had + 1/B and the second equation had - 1/B, adding the equations together eliminated B from the equation and allowed you to solve for A.\r<BR>\n" );
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document.write( "once you solved for A, it was a matter of replacing A with it's value to solve for B in either of the original equations.\r<BR>\n" );
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document.write( "here's a reference on how to solve simultaneous equations if you think you might need it.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.purplemath.com/modules/systlin1.htm\" target = \"_blank\">http://www.purplemath.com/modules/systlin1.htm</a>\r<BR>\n" );
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document.write( "here's a reference on how to add fractions with different denominators if you think you need it.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.purplemath.com/modules/fraction4.htm\" target = \"_blank\">http://www.purplemath.com/modules/fraction4.htm</a> </SPAN>\n" );
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