document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=427267>Question 427267</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi I really need help with this math problem----->                            find the LCM of 5(y-11)and 55(y-11) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #297131 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=297131\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i believe the answer is going to be 55*(y-11)\r<BR>\n" );
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document.write( "5*(y-11) * 11 = 55*(y-11)\r<BR>\n" );
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document.write( "55*(y-11) * 1 = 55*(Y-11)\r<BR>\n" );
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document.write( "look at the multiples of 5*(y-11)\r<BR>\n" );
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document.write( "they will be:\r<BR>\n" );
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document.write( "5*(y-11)<BR>\n" );
document.write( "10*(y-11)<BR>\n" );
document.write( "15*(y-11)<BR>\n" );
document.write( "20*(y-11)<BR>\n" );
document.write( "25*(y-11)<BR>\n" );
document.write( "30*(y-11)<BR>\n" );
document.write( "35*(y-11)<BR>\n" );
document.write( "40*(y-11)<BR>\n" );
document.write( "45*(y-11)<BR>\n" );
document.write( "50*(y-11)<BR>\n" );
document.write( "55*(y-11) *****<BR>\n" );
document.write( "etc.\r<BR>\n" );
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document.write( "the multiples of 55*(y-11) are:\r<BR>\n" );
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document.write( "55*(y-11)<BR>\n" );
document.write( "1108(y-11) *****<BR>\n" );
document.write( "etc.\r<BR>\n" );
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document.write( "to get the multiples, you just keep multiplying the original number by 1, then 2, then 3, then 4, etc.\r<BR>\n" );
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document.write( "alternatively, you just keep adding the number to itself.\r<BR>\n" );
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document.write( "the smallest multiple that can possibly be common to both would be 55 * (y-11)\r<BR>\n" );
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document.write( "5 * (y-11) * 11 = 55 * (y-11)\r<BR>\n" );
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document.write( "the smallest multiple therefore has to be 55 * (y-11)\r<BR>\n" );
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document.write( "you could have simplified the numbers together to make sure you were not doing anything wrong.\r<BR>\n" );
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document.write( "you would have gotten 5 * (y-11) = 5y - 55<BR>\n" );
document.write( "you would also have gotten 55 * (y-11) = 55y - 605\r<BR>\n" );
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document.write( "your multiples for 5y - 55 would have been:\r<BR>\n" );
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document.write( "5y-55<BR>\n" );
document.write( "10y-110<BR>\n" );
document.write( "15y-165<BR>\n" );
document.write( "20y-220<BR>\n" );
document.write( "25y-275<BR>\n" );
document.write( "30y-330<BR>\n" );
document.write( "35y-385<BR>\n" );
document.write( "40y-440<BR>\n" );
document.write( "45y-495<BR>\n" );
document.write( "50y-550<BR>\n" );
document.write( "55y-605 *****<BR>\n" );
document.write( "etc.\r<BR>\n" );
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document.write( "the common multiples for 55y-605 would have been:\r<BR>\n" );
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document.write( "55y-605 *****<BR>\n" );
document.write( "110y-1210<BR>\n" );
document.write( "etc.\r<BR>\n" );
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document.write( "you would get the same answer.\r<BR>\n" );
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document.write( "i believe your answer has to be that the least common multiple of 5*(y-11) and 55*(y-11) has to be 55*(y-11).\r<BR>\n" );
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document.write( "fyi:\r<BR>\n" );
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document.write( "the least common multiple has to be exactly divisible by each of the numbers involved.\r<BR>\n" );
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document.write( "in your problem the least common multiple is 55*(y-11)\r<BR>\n" );
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document.write( "that is exactly divisible by 5*(y-11).   it divides into it exactly 11 times.\r<BR>\n" );
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document.write( "that is also exactly divisible by 55*(y-11).   it divides into it exactly 1 time.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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