document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=845209>Question 845209</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A number is 6 greater than 1/2 another number.  The sum is 21, find the 2 numbers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #509176 by </B><A HREF=/tutors/aboutme.mpl?userid=pmesler><B>pmesler(52)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=pmesler><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=pmesler><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=509176\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = some number\r<BR>\n" );
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document.write( "Let y = the second number.\r<BR>\n" );
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document.write( "Therefore x = (y/2 + 6).\r<BR>\n" );
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document.write( "The problem says that x+y = 21.\r<BR>\n" );
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document.write( "All we have to do is substitute what we know x is into the equation and solve for y.\r<BR>\n" );
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document.write( "Therefore x+y = 21 becomes\r<BR>\n" );
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document.write( "(y/2 + 6) + y = 21\r<BR>\n" );
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document.write( "Now we need to isolate the y on the left side, but before we do that we need to combine like terms. We can't combine the y terms until both y's share the same denominator. To do this we need to multiply the second y by 2/2. That will give us the following:\r<BR>\n" );
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document.write( "(y/2 + 6) + y(2/2) = 21\r<BR>\n" );
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document.write( "We can multiply by 2/2 because 2/2 = 1 and multiplying anything by 1 does not change the value of the number. All we're doing is changing the form, not the value of the expression. \r<BR>\n" );
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document.write( "Our new equation is this:\r<BR>\n" );
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document.write( "(y/2 + 6) + (2y/2) = 21\r<BR>\n" );
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document.write( "Now we can combine the y terms and we get\r<BR>\n" );
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document.write( "3y/2 + 6 = 21.\r<BR>\n" );
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document.write( "Subtract 6 from both sides to simplify.\r<BR>\n" );
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document.write( "3y/2 = 15\r<BR>\n" );
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document.write( "Now to get y by itself we need to multitply both sides by the reciprocal of 3/2 which is 2/3. We do this so the numerator and denominators divide out and all we are left with is y.\r<BR>\n" );
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document.write( "(2/3)(3y/2) = 15(2/3).\r<BR>\n" );
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document.write( "y = 15 * 2/3\r<BR>\n" );
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document.write( "y = 30/3\r<BR>\n" );
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document.write( "y = 10. Now that we know that y = 10, it will be simple to find x.\r<BR>\n" );
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document.write( "x as you remember is (y/2 + 6)\r<BR>\n" );
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document.write( "Therefore x = (10/2 + 6)\r<BR>\n" );
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document.write( "x = 5 + 6\r<BR>\n" );
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document.write( "x = 11. \r<BR>\n" );
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document.write( "This checks out since 11 + 10 = 21.  </SPAN>\n" );
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