document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1077375>Question 1077375</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ava's mother is 3 times as old as Ava. Her grandmother is twice as old as Ava's mother. The sum of the numbers is 120. How old are Ava, her mother, and her grandmother?  I figured out the ages, but have no clue how to write the equation.   <BR>\n" );
document.write( "Ava is 12.<BR>\n" );
document.write( "Mother is 36.<BR>\n" );
document.write( "Grandmother is 72.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #691840 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=691840\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Start by assigning a variable to an unknown:\r<BR>\n" );
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document.write( "Let A = Ava's age\r<BR>\n" );
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document.write( "Then express other unknowns as a relation to this one (or to other assigned variables).  For each unknown, you will need to find an independent (meaning not just a previously found equation multiplied by a constant) equation. \r<BR>\n" );
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document.write( "Let M = Ava's mother's age<BR>\n" );
document.write( "We are given M=3A  (eq 1)\r<BR>\n" );
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document.write( "Let G = Ava's grandmother's age<BR>\n" );
document.write( "We are given G=2M   (eq 2)\r<BR>\n" );
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document.write( "The 3rd equation comes from the given information about the sum of their ages totaling 120:<BR>\n" );
document.write( "A+M+G = 120   (eq 3)\r<BR>\n" );
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document.write( "Now, (eq 3) can be written in terms of A alone, by substituting the values from (eq 1) and (eq 2).  First though we need to write (eq 2) in terms of A alone:\r<BR>\n" );
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document.write( "  G=2M, but M=3A, so G=2(3A) = 6A  (eq 2')\r<BR>\n" );
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document.write( "Now, substituting into (eq 3):<BR>\n" );
document.write( " A + M + G = 120 <BR>\n" );
document.write( " A + (3A) + (6A) = 120<BR>\n" );
document.write( "  10A =  120<BR>\n" );
document.write( "     A = 120/10 = 12  —>  M=36 —> G=72\r<BR>\n" );
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document.write( "—<BR>\n" );
document.write( "I hope this helps you understand the steps.   There are alternate methods to solving, but this problem lends itself to substitution.  <BR>\n" );
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