document.write( "Question 48491This question is from textbook College Algebra
\n" ); document.write( ": The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination.\r
\n" ); document.write( "\n" ); document.write( "x + y + 6z = 3
\n" ); document.write( "x + y + 3z = 3
\n" ); document.write( "x + 2y+ 4z = 7\r
\n" ); document.write( "\n" ); document.write( "I know this has to do with combining rows and eliminating variables until only one is left, then back-substituting into the system to find the remaining variables, but I honestly don't know where to begin the elimination.\r
\n" ); document.write( "\n" ); document.write( "Thank you so very much for all your help! \n" ); document.write( "
Algebra.Com's Answer #32086 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
x +  y + 6z = 3\r\n" );
document.write( "x +  y + 3z = 3\r\n" );
document.write( "x + 2y + 4z = 7\r\n" );
document.write( "\r\n" );
document.write( "I will assume you have had matrices. If not,\r\n" );
document.write( "post again and I will show you how to solve\r\n" );
document.write( "it without matrices:\r\n" );
document.write( "\r\n" );
document.write( "The augmented matrix is\r\n" );
document.write( "\r\n" );
document.write( "[1   1   6  |  3]\r\n" );
document.write( "[1   1   3  |  3]\r\n" );
document.write( "[1   2   4  |  7]\r\n" );
document.write( "\r\n" );
document.write( "We need to get 0's in the three lower left\r\n" );
document.write( "positions, that is, in the positions below\r\n" );
document.write( "the upper left to lower-right diagonal:\r\n" );
document.write( "\r\n" );
document.write( "To get a 0 where the 1 is in row 2 column 1 is,\r\n" );
document.write( "we multiply row 1 temporarily by -1 and add it\r\n" );
document.write( "to 1 times row 2. This is easy to do mentally\r\n" );
document.write( "if you write -1 to the left of row 1 and 1 left\r\n" );
document.write( "of row 2:\r\n" );
document.write( "\r\n" );
document.write( "-1[1   1   6  |  3]\r\n" );
document.write( " 1[1   1   3  |  3]\r\n" );
document.write( "  [1   2   4  |  7]\r\n" );
document.write( "\r\n" );
document.write( "  [1   1   6  |  3]\r\n" );
document.write( "  [0   0  -3  |  0]\r\n" );
document.write( "  [1   2   4  |  7]\r\n" );
document.write( "\r\n" );
document.write( "To get a 0 where the 1 is in row 3 column 1 is,\r\n" );
document.write( "we multiply row 1 temporarily by -1 and add it\r\n" );
document.write( "to 1 times row 3. This is easy to do mentally\r\n" );
document.write( "if you write -1 to the left of row 1 and 1 left\r\n" );
document.write( "of row 3:\r\n" );
document.write( "\r\n" );
document.write( " -1[1   1   6  |  3]\r\n" );
document.write( "   [0   0  -3  |  0]\r\n" );
document.write( "  1[1   2   4  |  7]\r\n" );
document.write( "\r\n" );
document.write( "   [1   1   6  |  3]\r\n" );
document.write( "   [0   0  -3  |  0]\r\n" );
document.write( "   [0   1  -2  |  4]\r\n" );
document.write( "\r\n" );
document.write( "To get a 0 where the 1 is in row 3 column 2 is,\r\n" );
document.write( "we merely need to swap rows 2 and 3\r\n" );
document.write( "\r\n" );
document.write( "   [1   1   6  |  3]\r\n" );
document.write( "   [0   1  -2  |  4]\r\n" );
document.write( "   [0   0  -3  |  0]\r\n" );
document.write( " \r\n" );
document.write( "Now we need to get 1's on the upper left to\r\n" );
document.write( "lower right diagonal.  Two of the diagonal \r\n" );
document.write( "elements are already 1's.  To get a 1 where \r\n" );
document.write( "the -3 is, we multiply row 3 by -1/3\r\n" );
document.write( "\r\n" );
document.write( "    [1   1   6  |  3]\r\n" );
document.write( "    [0   1  -2  |  4]\r\n" );
document.write( "-1/3[0   0  -3  |  0]\r\n" );
document.write( "\r\n" );
document.write( "   [1   1   6  |  3]\r\n" );
document.write( "   [0   1  -2  |  4]\r\n" );
document.write( "   [0   0   1  |  0]\r\n" );
document.write( "\r\n" );
document.write( "This is the system:\r\n" );
document.write( "\r\n" );
document.write( "   1x + 1y + 6z = 3\r\n" );
document.write( "   0x + 1y - 2z = 4\r\n" );
document.write( "   0x + 0y + 1z = 0\r\n" );
document.write( "\r\n" );
document.write( "or\r\n" );
document.write( "\r\n" );
document.write( "    x +  y + 6z = 3\r\n" );
document.write( "         y - 2z = 4\r\n" );
document.write( "              z = 0\r\n" );
document.write( "\r\n" );
document.write( "The bottom equation tells us that\r\n" );
document.write( "z = 0\r\n" );
document.write( "\r\n" );
document.write( "Substitute z = 0 into the 2nd equation\r\n" );
document.write( "\r\n" );
document.write( "      x + y + 6z = 3\r\n" );
document.write( "          y - 2z = 4\r\n" );
document.write( "         y - 2·0 = 4\r\n" );
document.write( "               y = 4\r\n" );
document.write( "\r\n" );
document.write( "Substitute z = 0 and y = 4 into the 1st\r\n" );
document.write( "equation\r\n" );
document.write( "\r\n" );
document.write( "     x + 4 + 6·0 =  3\r\n" );
document.write( "       x + 4 + 0 =  3\r\n" );
document.write( "               x = -1\r\n" );
document.write( "\r\n" );
document.write( "So the solution is (x, y, z) = (-1, 4, 0)\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );