document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=139405>Question 139405</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0078651131/algebrahomeworkh>Algebra 1</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I need to solve this equation and state any extraneous solutions. When it says over, it's a fraction. I didn't know how else to put it.<BR>\n" );
document.write( "a over 3a+6<BR>\n" );
document.write( "minus<BR>\n" );
document.write( "a over 5a+10<BR>\n" );
document.write( "equals 2 over 5\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #101655 by </B><A HREF=/tutors/aboutme.mpl?userid=frostusna><B>frostusna(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=frostusna><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=101655\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The equation form of your problem would look like this:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2F%283a%2B6%29-+a%2F%285a%2B10%29+=+2%2F5 ALIGN=MIDDLE  ALT=\"a%2F%283a%2B6%29-+a%2F%285a%2B10%29+=+2%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Since the two terms on the left hand side have different denominators they can't be subtracted as they are, therefor we must give them common denominators. We do that by multiplying each term's numerator and denominator by the denominator of the other term, esentially multiplying each term by the number 1. The equation becomes:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%285a%2B10%29%2Aa%29%2F%285a%2B10%29%283a%2B6%29+-+%28%283a%2B6%29%2Aa%29%2F%283a%2B6%29%285a%2B10%29+=+2%2F5 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Multiplying out the numerators and denominators renders:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%285a%5E2+%2B+10a%29+-+%283a%5E2+%2B+6a%29%29%2F%2815a%5E2+%2B+60a+%2B+60%29+=+2%2F5 ALIGN=MIDDLE  ALT=\"%28%285a%5E2+%2B+10a%29+-+%283a%5E2+%2B+6a%29%29%2F%2815a%5E2+%2B+60a+%2B+60%29+=+2%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now combining like terms in the numerator and multiplying each side by the denominator of the left hand term gives:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282a%5E2+%2B+4a%29+=+%282%2F5%29+%2A+%2815a%5E2+%2B+60a+%2B+60%29 ALIGN=MIDDLE  ALT=\"%282a%5E2+%2B+4a%29+=+%282%2F5%29+%2A+%2815a%5E2+%2B+60a+%2B+60%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Or: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282a%5E2+%2B+4a%29+=+6a%5E2+%2B+24a+%2B+24 ALIGN=MIDDLE  ALT=\"%282a%5E2+%2B+4a%29+=+6a%5E2+%2B+24a+%2B+24\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Now, subtract <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2a%5E2+%2B+4a ALIGN=MIDDLE  ALT=\"2a%5E2+%2B+4a\"   DISABLED_style= \"max-width:90%; height:auto;\" > from each side and we end up with the quadratic equation:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4a%5E2+%2B+20a+%2B+24+=+0 ALIGN=MIDDLE  ALT=\"4a%5E2+%2B+20a+%2B+24+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Then, using the quadratic equation solver we get:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a+=+%28-20%2B-sqrt%2820%5E2-4%2A4%2A24%29%29%2F%282%2A4%29 ALIGN=MIDDLE  ALT=\"a+=+%28-20%2B-sqrt%2820%5E2-4%2A4%2A24%29%29%2F%282%2A4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "Solving this equation yields two solutions:\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=-2 ALIGN=MIDDLE  ALT=\"a=-2\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=-3 ALIGN=MIDDLE  ALT=\"a=-3\"   DISABLED_style= \"max-width:90%; height:auto;\" >. However, when the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=-2 ALIGN=MIDDLE  ALT=\"a=-2\"   DISABLED_style= \"max-width:90%; height:auto;\" > is substituted into the original equation you end up with 0 (zero) in both denominators, which makes the terms undefinable. So the one workable solution is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=-3 ALIGN=MIDDLE  ALT=\"a=-3\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "\n" );
document.write( "I hope this helped. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );