document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=91629>Question 91629</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I don't understand how to solve this complicated linear equation word problem. I would really appreciate if you could help me out, thanks.\r<BR>\n" );
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document.write( "THE SUM OF TWO NUMBERS IS 36. IF THE LARGER NUMBER IS DIVIDED BY THE SMALLER NUMBER, THE QUOTIENT IS 2 AND THE REMAINDER IS 3. FIND THE NUMBERS.\r<BR>\n" );
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document.write( "I'm sorry but this isn't a textbook question </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #67335 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=67335\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's begin by calling the smaller number S and the larger number L. <BR>\n" );
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document.write( "The problem tells you that the sum of these two numbers is 36. Therefore, we can write the<BR>\n" );
document.write( "equation:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=S+%2B+L+=+36 ALIGN=MIDDLE  ALT=\"S+%2B+L+=+36\"   DISABLED_style= \"max-width:90%; height:auto;\" > <=== first equation<BR>\n" );
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document.write( "The the problem tells you to divide the larger number by the smaller number. In algebraic<BR>\n" );
document.write( "form this would be <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L%2FS ALIGN=MIDDLE  ALT=\"L%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Furthermore it tells you that the result of this division equals 2 plus a remainder of<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2FS ALIGN=MIDDLE  ALT=\"3%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "So the equation form of this division is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L%2FS+=+2+%2B+3%2FS ALIGN=MIDDLE  ALT=\"L%2FS+=+2+%2B+3%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" > <=== second equation<BR>\n" );
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document.write( "If we return to the first equation and subtract S from both sides, we get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L+=+36+-+S ALIGN=MIDDLE  ALT=\"L+=+36+-+S\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Since L equals 36 - S we can substitute 36 - S for L in the second equation which then<BR>\n" );
document.write( "becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2836+-+S%29%2FS+=+2+%2B+3%2FS ALIGN=MIDDLE  ALT=\"%2836+-+S%29%2FS+=+2+%2B+3%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Notice that if we say the first term on the right side (that term is 2) is equivalent to<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2S%2FS ALIGN=MIDDLE  ALT=\"2S%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" > then we can substitute <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2S%2FS ALIGN=MIDDLE  ALT=\"2S%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" > for 2 and every term in the equation will <BR>\n" );
document.write( "then have an S in the denominator and the equation will be:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2836+-+S%29%2FS+=+2S%2FS+%2B+3%2FS ALIGN=MIDDLE  ALT=\"%2836+-+S%29%2FS+=+2S%2FS+%2B+3%2FS\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "We can now get rid of the S in the denominator by multiplying all terms on both sides by S<BR>\n" );
document.write( "to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=36+-+S+=+2S+%2B+3 ALIGN=MIDDLE  ALT=\"36+-+S+=+2S+%2B+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now add S to both sides of the equation to get rid of the -S on the left side. The equation<BR>\n" );
document.write( "then becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=36+=+3S+%2B+3 ALIGN=MIDDLE  ALT=\"36+=+3S+%2B+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Then subtract 3 from both sides to get rid of the 3 on the right side. The equation<BR>\n" );
document.write( "then becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=33+=+3S ALIGN=MIDDLE  ALT=\"33+=+3S\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Finally, divide both sides by 3 to solve for S:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=S+=+33%2F3+=+11 ALIGN=MIDDLE  ALT=\"S+=+33%2F3+=+11\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "So now we know the smaller number is 11. But since the sum of the two numbers is 36, then<BR>\n" );
document.write( "the larger number must be 25 because 36 - 11 = 25.<BR>\n" );
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document.write( "So the two numbers are 11 and 25.<BR>\n" );
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document.write( "Check: is their sum 36?  Yes it is.<BR>\n" );
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document.write( "If you divide the larger (25) by the smaller (11) do you get an answer of 2 with a remainder<BR>\n" );
document.write( "of 3? Yes you do.  Our answer checks ... the smaller number is 11 and the larger number is 25.<BR>\n" );
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document.write( "Hope this helps you to see your way through the problem.<BR>\n" );
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document.write( " </SPAN>\n" );
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