document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=755962>Question 755962</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how can i solve this system of equations using the addition method when one or more of the terms is missing?<BR>\n" );
document.write( "4a+9b=8<BR>\n" );
document.write( "8a+6c=-1<BR>\n" );
document.write( "6b+6c=-1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #459923 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=459923\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> your original equations are:\r<BR>\n" );
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document.write( "4a+9b=8 (equation 1)<BR>\n" );
document.write( "8a+6c=-1 (equation 2)<BR>\n" );
document.write( "6b+6c=-1 (equation 3)\r<BR>\n" );
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document.write( "fill in the missing variables with 0 times that variable to get:\r<BR>\n" );
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document.write( "4a + 9b + 0c = 8  (equation 1)<BR>\n" );
document.write( "8a + 0b + 6c = 1  (equation 2)<BR>\n" );
document.write( "0a + 6b + 6c = 1  (equation 3)\r<BR>\n" );
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document.write( "now you have 3 equations in 3 unknowns which can be solved.\r<BR>\n" );
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document.write( "work on the first 2 equations and eliminate the a variable by multiplying the first equation by 2 and then subtracting the second equation from it.\r<BR>\n" );
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document.write( "after you multiply equation 1 by 2, it becomes equation 4.\r<BR>\n" );
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document.write( "you get:\r<BR>\n" );
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document.write( "8a + 18b + 0c = 16 (equation 4)<BR>\n" );
document.write( "minus:<BR>\n" );
document.write( "8a + 0b + 6c = 1 (equation 2)<BR>\n" );
document.write( "equals:<BR>\n" );
document.write( "0a + 18b - 6c = 15 (equation 5)\r<BR>\n" );
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document.write( "the result of the subtraction becomes equation 5.\r<BR>\n" );
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document.write( "since the a variable is already eliminated from equation 3, you don't have anything to do except take the remaining 2 equations (equation 5 and equation 3) and solve them as 2 equations in 2 unknowns.\r<BR>\n" );
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document.write( "you have:\r<BR>\n" );
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document.write( "18b - 6c = 15 (equation 5)<BR>\n" );
document.write( "6b + 6c = 1 (equation 3)\r<BR>\n" );
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document.write( "since you have + 6c in equation 3 and - 6c in equation 5, all you have to do is add these equations together and you will have eliminated the c variable which will allow you to solve for b.\r<BR>\n" );
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document.write( "you get:\r<BR>\n" );
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document.write( "18b - 6c = 15 (equation 5)<BR>\n" );
document.write( "plus:<BR>\n" );
document.write( "6b + 6c = 1 (equation 3)<BR>\n" );
document.write( "equals:<BR>\n" );
document.write( "24b = 16 \r<BR>\n" );
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document.write( "solve for b to get:\r<BR>\n" );
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document.write( "b = 2/3\r<BR>\n" );
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document.write( "you now have the value of b which you can now use to solve for c in equations 5 or 3.\r<BR>\n" );
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document.write( "either one will do, and you should get the same answer regardless of which one you choose.\r<BR>\n" );
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document.write( "i'll do both to show you.\r<BR>\n" );
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document.write( "equations 5 and 3 are:\r<BR>\n" );
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document.write( "18b - 6c = 15 (equation 5)<BR>\n" );
document.write( "6b + 6c = 1 (equation 3)\r<BR>\n" );
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document.write( "replace b with 2/3 in each of these questions and solve for c to get:\r<BR>\n" );
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document.write( "c = -(1/2) in equation 5.<BR>\n" );
document.write( "c = -(1/2) in equation 3.\r<BR>\n" );
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document.write( "now you have:\r<BR>\n" );
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document.write( "b = 2/3<BR>\n" );
document.write( "c = -(1/2)\r<BR>\n" );
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document.write( "go back to your original 3 equations and use the values of b and c to solve for a in any one of those equations that has a in it.\r<BR>\n" );
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document.write( "you only need to solve for a in one of those equation and then you can confirm by using the values of a,b,c in all 3 equations.  if the equations are true, then you did good.\r<BR>\n" );
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document.write( "equation 1 states that 4a + 9b = 8<BR>\n" );
document.write( "if b = 2/3, then this equation becomes:<BR>\n" );
document.write( "4a + 9(2/3) = 8 which becomes:<BR>\n" );
document.write( "4a + 6 = 8 which becomes:<BR>\n" );
document.write( "4a = 2 which becomes:<BR>\n" );
document.write( "a = (1/2)\r<BR>\n" );
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document.write( "you now have:\r<BR>\n" );
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document.write( "a = 1/2<BR>\n" );
document.write( "b = 2/3<BR>\n" );
document.write( "c = -(1/2)\r<BR>\n" );
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document.write( "plug these values in all 3 original equations and they should all be true.<BR>\n" );
document.write( "if so, you're done.<BR>\n" );
document.write( "if not, you messed up somewhere and have to go back and see where you went wrong.\r<BR>\n" );
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document.write( "the original 3 equations are:\r<BR>\n" );
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document.write( "4a + 9b = 8  (equation 1)<BR>\n" );
document.write( "8a + 6c = 1  (equation 2)<BR>\n" );
document.write( "6b + 6c = 1  (equation 3)\r<BR>\n" );
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document.write( "replacing a with 1/2 and b with 2/3 and c with -(1/2), get you:\r<BR>\n" );
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document.write( "4(1/2) + 9(2/3) = 8<BR>\n" );
document.write( "8(1/2) + 6(-(1/2)) = 1<BR>\n" );
document.write( "6(2/3) + 6(-(1/2)) = 1\r<BR>\n" );
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document.write( "solving each of these equations gets:\r<BR>\n" );
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document.write( "8=8 (true)<BR>\n" );
document.write( "1=1 (true)<BR>\n" );
document.write( "1-1 (true)\r<BR>\n" );
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document.write( "you did good.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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