document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=164811>Question 164811</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the sum of the ages of two sisters is 37. the difference between four times the older sisters's age and three times the younger sistter's age is 43. how can i find the age of these sisters? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #121448 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=121448\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> lets call the younger sister s1 and the older sister s2. now we know that s1+s2=37 and we know that difference between 4 times the older sisters age which we will write as 4s2 and  3 times the younger sister age which we will write as 3s1 equals 43..writing this out this would be 4s2-3s1=43. So solving for s1 in the 1st equation we get s1=37-s2   plugging that into the second equation we get <BR>\n" );
document.write( "4s2-3(37-s2)=43.  now multiply out and simplify...we get 7s2=154 divide by 7 and s2=22        so since s1=37-s2(22) we get s1 is 15\r<BR>\n" );
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document.write( "answers younger sister 15<BR>\n" );
document.write( "        older   sister 22 </SPAN>\n" );
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