document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1203019>Question 1203019</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have a problem with questions relating to variation as the sum of two parts.<BR>\n" );
document.write( "Here is a worked example.<BR>\n" );
document.write( "A quantity, p, is the sum of two terms, one of which is constant while the other varies inversely as the square of q.<BR>\n" );
document.write( "When q = 1, p=-1;<BR>\n" );
document.write( "when q = 2, p=2;<BR>\n" );
document.write( "find the positive value of q when p=2.75<BR>\n" );
document.write( "If a and b are constants, p=a+b/q2;<BR>\n" );
document.write( "q = 1, p= -1<BR>\n" );
document.write( "-1 = a +b/12;<BR>\n" );
document.write( "therefore -1 = a+b (equation 1)<BR>\n" );
document.write( "when q = 2, p=2<BR>\n" );
document.write( "2 = a+b/22<BR>\n" );
document.write( "2=a+b/4<BR>\n" );
document.write( "8 = 4a+b (equation 2)<BR>\n" );
document.write( "Subtracting equation 1 from 2 gives<BR>\n" );
document.write( "9=3a<BR>\n" );
document.write( "so a =3<BR>\n" );
document.write( "Substituting a = 3 into equation 1 we get<BR>\n" );
document.write( "-1 = 3+b<BR>\n" );
document.write( "b=-4<BR>\n" );
document.write( "so p=3-4/q2<BR>\n" );
document.write( "so when p = 2.75, 2.75 = 3-4/q2<BR>\n" );
document.write( "4/q2 = ¼<BR>\n" );
document.write( "q2 = 16<BR>\n" );
document.write( "q=+/- 4 and we want the positive value so q = 4<BR>\n" );
document.write( "I have tried to apply the same approach to the following problem:<BR>\n" );
document.write( "A quantity C is the sum of two parts.  The first part varies directly as the cube of t;<BR>\n" );
document.write( "the second part varies inversely as the square of t.<BR>\n" );
document.write( "C = 74, t = 1<BR>\n" );
document.write( "C = 34, t = 2<BR>\n" );
document.write( "Find C when t = 3<BR>\n" );
document.write( "I wrote C = a3 +b/t2<BR>\n" );
document.write( "74 = a3+b/12<BR>\n" );
document.write( "74 = a3 +b (equation 1).\r<BR>\n" );
document.write( "\n" );
document.write( "34 = a3+b/22<BR>\n" );
document.write( "34 = a3+b/4<BR>\n" );
document.write( "136 = 4a3+b (equation 2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Subtract equation 1 from 2<BR>\n" );
document.write( "136 = 4a3+b<BR>\n" );
document.write( "74 = a3 +b<BR>\n" );
document.write( "This gives 62 = 3a3<BR>\n" );
document.write( "a3 = 20.67<BR>\n" );
document.write( "a = 2.74\r<BR>\n" );
document.write( "\n" );
document.write( "Substituting into the first equation<BR>\n" );
document.write( "74 = 20.67 + b/12<BR>\n" );
document.write( "74 = 20.67+b<BR>\n" );
document.write( "b=53.33\r<BR>\n" );
document.write( "\n" );
document.write( "To find C when t = 3<BR>\n" );
document.write( "C = a3+b/t2<BR>\n" );
document.write( "C = 20.67 + 53.33/9<BR>\n" );
document.write( "9C = 186.03+53.33<BR>\n" );
document.write( "9C = 239.36<BR>\n" );
document.write( "C = 25.6 but the answer is 62.<BR>\n" );
document.write( "Where have I gone wrong?\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #838250 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=838250\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> your problem seems to be that you assumed that the constant of variation was the same in both parts.<BR>\n" );
document.write( "it is not.<BR>\n" );
document.write( "this apparently didn't apply to the first problem you solved because one of the parts was a constant and didn't involve a constant of variation.<BR>\n" );
document.write( "in the second problem, both parts required a constant of variation.<BR>\n" );
document.write( "i initially assumed both constants of variation would be the same but ran into trouble<BR>\n" );
document.write( "i then went back and assumed that they were different for each part.<BR>\n" );
document.write( "this is how i solved it.<BR>\n" );
document.write( "you are given that C = a + b, where a = x * t^3 and b = y / t^2.<BR>\n" );
document.write( "both x and y are constants of variations for their part.<BR>\n" );
document.write( "you are then given that C = 74 when t = 1 and C = 34 when t = 2<BR>\n" );
document.write( "your two equations become:<BR>\n" );
document.write( "74 = x * 1^3 + y / 1^2 when t = 1<BR>\n" );
document.write( "34 = x * 2^3 + y / 2^2 when t = 2<BR>\n" );
document.write( "the first equation becomes 74 = x + y<BR>\n" );
document.write( "the second equation bcomes 34 = 8x + y/4<BR>\n" );
document.write( "you now want to solve these equations simultaneously.<BR>\n" );
document.write( "first equation becomes x + y = 74<BR>\n" );
document.write( "second equation becomes 32x + y = 136<BR>\n" );
document.write( "subtract the first equation from the second to get 31x = 62<BR>\n" );
document.write( "solve for x to get x = 2<BR>\n" );
document.write( "since x + y = 74, then y must be equal to 72.<BR>\n" );
document.write( "you now have x = 2 and y = 72<BR>\n" );
document.write( "x is the constant of variation for the first part and y is the constant of vriation for the second part.<BR>\n" );
document.write( "they want you to find C when t = 3.<BR>\n" );
document.write( "when t = 3, your equation becomes C = 2 * 3^3 = 72 / 3^2.<BR>\n" );
document.write( "this becomes C = 2 * 27 + 72 / 9<BR>\n" );
document.write( "this becomes C = 54 + 8 which becomes C = 62.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "i'm pretty sure i did it right.<BR>\n" );
document.write( "your biggest problem appears that you assumed the constant of variation was the same in both parts.<BR>\n" );
document.write( "that's what i think.<BR>\n" );
document.write( "i also think think that you assumed b was the constant of variation in the first equation.<BR>\n" );
document.write( "it was not.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );