document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=67624>Question 67624</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have always had problems with fractions. If you could help me out so I can do this in the future, that would be great.\r<BR>\n" );
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document.write( "1/8 + 5/6 - 2/3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #48120 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=48120\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> FIND A COMMON DENOMINATOR . THAT IS A NUMBER DIVISABLE BY 8,6&3.<BR>\n" );
document.write( "AN EASY METHOD IS TO MULTIPLY ALL THREE 8*6*3=144.<BR>\n" );
document.write( "HOWEVER A SMALLER COMMON DENOMINATOR CAN BE FOUND BY FACTORING ALL NUMBERS THUS:<BR>\n" );
document.write( "8=(2)*2*2<BR>\n" );
document.write( "6=(2)*(3)<BR>\n" );
document.write( "3=(3) COMMOM FACTORS ARE (3) & ONE SET OF (2) SO WE ONLY NEED TO USE THESE DUPLICATE FACTORS ONCE THUS 2*3. THEN THE REMAINING FACTORS 2*2<BR>\n" );
document.write( "WE NOW HAVE 2*3*2*2=24 THIS IS THE LOWEST COMMON DENOMINATOR.<BR>\n" );
document.write( "1/8+5/6-2/3 NOW DIVIDE 24 BY EACH DENOMINATOR & MULTIPLY THIS NUMBER BY THE NUMERATOR THEN ADD ALL NUMERATORS OVER A DENOMINATOR OF 24:<BR>\n" );
document.write( "(1*3)+(5*4)-(2*8)/24<BR>\n" );
document.write( "(3+20-16)/24<BR>\n" );
document.write( "7/24<BR>\n" );
document.write( "HAPPY CALCULATING  : )<BR>\n" );
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