document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=116081>Question 116081</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If you could explain to me how to solve this problem, I would appreciate it. I need numerical values for a, b, c, and d. I know this involves substitution, I just don't know where to start! Thank you so much. \r<BR>\n" );
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document.write( "2a+3b-4c+6d=6\r<BR>\n" );
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document.write( "7a-5b-c=-7\r<BR>\n" );
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document.write( "13a-9b=6\r<BR>\n" );
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document.write( "d^2-2d=-1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #84413 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=84413\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Sort of complex. For reference I'm going to use Roman Numerals to Identify the 4 equations as follows:<BR>\n" );
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document.write( "(I) 2a+3b-4c+6d=6<BR>\n" );
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document.write( "(II) 7a-5b-c=-7<BR>\n" );
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document.write( "(III) 13a-9b=6<BR>\n" );
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document.write( "(IV) d^2-2d=-1<BR>\n" );
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document.write( "Cramer's rule might be an interesting way to solve this, but I'm going to assume that you<BR>\n" );
document.write( "haven't studied that yet. So let's just plod our way through.<BR>\n" );
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document.write( "The first thing that I noticed was that equation (IV) has just one variable. So we can solve<BR>\n" );
document.write( "it for d first thing. Add 1 to both sides of equation (IV) and it becomes:<BR>\n" );
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document.write( "d^2 - 2d + 1 = 0<BR>\n" );
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document.write( "The left side is a perfect square as follows:<BR>\n" );
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document.write( "(d - 1)^2 = 0<BR>\n" );
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document.write( "To make the left side equal to the zero on the right side we need to have:<BR>\n" );
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document.write( "d - 1 = 0<BR>\n" );
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document.write( "Solve for d by adding 1 to both sides to get:<BR>\n" );
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document.write( "d = 1<BR>\n" );
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document.write( "One down ... only a, b, and c to find. Before we go further let's go back to equation (I) and <BR>\n" );
document.write( "substitute 1 for d to make that equation:<BR>\n" );
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document.write( "2a + 3b - 4c + 6(1) = 6<BR>\n" );
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document.write( "subtract 6 from both sides and equation I reduces to:<BR>\n" );
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document.write( "(I) 2a + 3b - 4c = 0<BR>\n" );
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document.write( "Notice that the form of this equation is now very similar to equation (II). Let's multiply<BR>\n" );
document.write( "both sides of equation (II) by -4 and it becomes:<BR>\n" );
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document.write( "(II) -28a + 20b + 4c = 28<BR>\n" );
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document.write( "We now have the new equations (I) and (II) as:<BR>\n" );
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document.write( "( I ) 2a + 3b - 4c = 0<BR>\n" );
document.write( "(II)-28a +20b + 4c = 28<BR>\n" );
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document.write( "Notice now what happens if we add these two equations vertically in columns. We eliminate<BR>\n" );
document.write( "the terms containing the variable c and the equation resulting from this addition (call<BR>\n" );
document.write( "it equation (I&II) is:<BR>\n" );
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document.write( "(I&II) -26a + 23b = 28<BR>\n" );
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document.write( "This looks a lot like equation (III) and the pair of equations is now:<BR>\n" );
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document.write( "(I&II)-26a + 23b = 28<BR>\n" );
document.write( "(III) +13a - 9b = 6<BR>\n" );
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document.write( "Let's work to eliminate the variable \"a\". Multiply equation (III) ... all terms on both<BR>\n" );
document.write( "sides by 2 to make the equation pair become:<BR>\n" );
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document.write( "(I&II)-26a + 23b = 28<BR>\n" );
document.write( "(III) +26a - 18b = 12<BR>\n" );
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document.write( "If we then add the two equations vertically in columns, the two \"a\" terms cancel each other<BR>\n" );
document.write( "and the combined equation that results from the addition (call it equation (I&II&III) becomes:<BR>\n" );
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document.write( "(I&II&III) 5b = 40<BR>\n" );
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document.write( "Divide both sides by 5 and get:<BR>\n" );
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document.write( "b = 40/5 = 8<BR>\n" );
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document.write( "Two down ... only a and c left to find.<BR>\n" );
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document.write( "Now that we know b = 8 we can return to the original equation (III) and substitute<BR>\n" );
document.write( "8 for b to get:<BR>\n" );
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document.write( "(III) 13a - 9(8) = 6<BR>\n" );
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document.write( "Multiply 9 times 8 and the equation changes to:<BR>\n" );
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document.write( "(III) 13a - 72 = 6<BR>\n" );
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document.write( "Add 72 to both sides and it further becomes:<BR>\n" );
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document.write( "(III) 13a = 78<BR>\n" );
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document.write( "Solve for a by dividing both sides by 13 to get:<BR>\n" );
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document.write( "a = 78/13 = 6<BR>\n" );
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document.write( "One more to go ... just c left.<BR>\n" );
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document.write( "Return to one of the original equations that contains c and substitute the known values for<BR>\n" );
document.write( "a and b. Let's go way back to equation (II). If we substitute 6 for a and 8 for b that equation<BR>\n" );
document.write( "becomes:<BR>\n" );
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document.write( "(II) 7(6)- 5(8) - c = -7<BR>\n" );
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document.write( "Do the multiplications and we get:<BR>\n" );
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document.write( "(II) 42 - 40 - c = -7<BR>\n" );
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document.write( "Combine the numbers on the left side:<BR>\n" );
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document.write( "(II) 2 - c = -7<BR>\n" );
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document.write( "Subtract 2 from both sides to reduce this to:<BR>\n" );
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document.write( "(II) - c = -9<BR>\n" );
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document.write( "Solve for +c by multiplying both sides by -1 and we have:<BR>\n" );
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document.write( "c = +9<BR>\n" );
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document.write( "That's it ... no more variables to find. In summary: a = 6, b = 8, c = 9, and d = 1.<BR>\n" );
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document.write( "You can check these answers by returning to the original 4 equations and substituting<BR>\n" );
document.write( "the values for a, b, c, and d as needed in each equation. You should (and will) find that<BR>\n" );
document.write( "with these values the left side of each equation will equal the right side.<BR>\n" );
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document.write( "Hope this helps you to understand the problem.<BR>\n" );
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