document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=491321>Question 491321</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do I solve this inequality? Please explain with detail, for I do not know how to solve such a problem. \r<BR>\n" );
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document.write( "x + x-3  > 6<BR>\n" );
document.write( "__  ___<BR>\n" );
document.write( "4    3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #334592 by </B><A HREF=/tutors/aboutme.mpl?userid=deborabr><B>deborabr(181)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=deborabr><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=deborabr><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=334592\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x/4 + (x-3)/3 > 6<BR>\n" );
document.write( "make the least common multiple of 4 and 3<BR>\n" );
document.write( "just divide the numbers together by primes numbers and multiply the divisors<BR>\n" );
document.write( "4,3|2<BR>\n" );
document.write( "2,3|2<BR>\n" );
document.write( "1,3|3<BR>\n" );
document.write( "1,1<BR>\n" );
document.write( "2*2*3=12<BR>\n" );
document.write( "12 is the least common multiple of 4 and 3<BR>\n" );
document.write( "now all the numbers have the same denominator but you need divide 12 by denominator of the equation and later multiply the result by the numerator<BR>\n" );
document.write( "for example 12/3 = 4 and 4*x = 4x so the firs is 3x/12 now make like this with the others of the equation<BR>\n" );
document.write( "making all we have:\r<BR>\n" );
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document.write( "3x/12 + 4(x-3)/12 > 72/12<BR>\n" );
document.write( "if all the denominators are equal we can delete them<BR>\n" );
document.write( "and stay like this<BR>\n" );
document.write( "3x + 4(x-3) > 72<BR>\n" );
document.write( "in 4(x-3) the 4 is multiplying x and -3 so 4*x=4x and 4*-3=-12<BR>\n" );
document.write( "3x + 4(x-3) > 72<BR>\n" );
document.write( "3x + 4x - 12 > 72<BR>\n" );
document.write( "3x + 4x > 72 + 12 single numbers with single numbers and numbers with \"x\" with numbers with \"x\"<BR>\n" );
document.write( "calculating<BR>\n" );
document.write( "7x > 84 the number that is multiplying \"x\" pass dividing<BR>\n" );
document.write( "x > 84/7<BR>\n" );
document.write( "x > 12 </SPAN>\n" );
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