document.write( "Question 86805: linear systems in three variables\r
\n" ); document.write( "\n" ); document.write( "2x+3y-6z=4
\n" ); document.write( "3x-2y-9z=-7
\n" ); document.write( "2x+5y-6z=8 \n" ); document.write( "
Algebra.Com's Answer #62980 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "linear systems in three variables \r\n" );
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document.write( "2x + 3y - 6z =  4\r\n" );
document.write( "3x - 2y - 9z = -7\r\n" );
document.write( "2x + 5y - 6z =  8\r\n" );
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document.write( "Make this augmented matrix of coefficients:\r\n" );
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document.write( "[2  3 -6 |  4]\r\n" );
document.write( "[3 -2 -9 | -7]\r\n" );
document.write( "[2  5 -6 |  8]\r\n" );
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document.write( "The idea is to get 0's where the three red\r\n" );
document.write( "numbers are.\r\n" );
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document.write( "[2  3 -6 |  4]\r\n" );
document.write( "[3 -2 -9 | -7]\r\n" );
document.write( "[2  5 -6 |  8]\r\n" );
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document.write( "[2  3 -6 |  4]\r\n" );
document.write( "[3 -2 -9 | -7]\r\n" );
document.write( "[2  5 -6 |  8]\r\n" );
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document.write( "Get a 0 where the red 3 is by multiplying row 1 by\r\n" );
document.write( "-3 and adding it to 2 times row 2, but restore row 1:\r\n" );
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document.write( "-3×[2  3 -6 |  4]\r\n" );
document.write( " 2×[3 -2 -9 | -7]\r\n" );
document.write( "   [2  5 -6 |  8]\r\n" );
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document.write( "[2   3  -6 |   4]\r\n" );
document.write( "[0 -13   0 | -26]\r\n" );
document.write( "[2   5  -6 |   8] \r\n" );
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document.write( "Get a 0 where the red 2 is by multiplying row 1 by\r\n" );
document.write( "-1 and adding it to 1 times row 3, but restore row 1:\r\n" );
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document.write( "-1×[2   3  -6 |   4]\r\n" );
document.write( "   [0 -13   0 | -26]\r\n" );
document.write( " 1×[2   5  -6 |   8]\r\n" );
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document.write( "[2   3  -6 |   4]\r\n" );
document.write( "[0 -13   0 | -26]\r\n" );
document.write( "[0   2   0 |   4]\r\n" );
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document.write( "Get a 0 where the red 2 is by multiplying row 2 by\r\n" );
document.write( "2 and adding it to 13 times row 3, but restore row 2:\r\n" );
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document.write( "   [2   3  -6 |   4]\r\n" );
document.write( " 2×[0 -13   0 | -26]\r\n" );
document.write( "13×[0   2   0 |   4] \r\n" );
document.write( "  \r\n" );
document.write( "[2   3  -6 |   4]\r\n" );
document.write( "[0 -13   0 | -26]\r\n" );
document.write( "[0   0   0 |   0]\r\n" );
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document.write( "Get a 1 where the red 2 is by dividing row 1 thru by 2\r\n" );
document.write( "Get a 1 where the red -13 is by dividing row 2 thru by -13\r\n" );
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document.write( "[1  \"3%2F2\" -3 | 2]\r\n" );
document.write( "[0   1  0 | 2]\r\n" );
document.write( "[0   0  0 | 0]\r\n" );
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document.write( "Convert back into a system of equations:\r\n" );
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document.write( "1x +  \"3%2F2\"y - 3z = 2\r\n" );
document.write( "0x +   1y + 0z = 2\r\n" );
document.write( "0x +   0y + 0z = 0\r\n" );
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document.write( "Erase the first 0 term of the 2nd equation,\r\n" );
document.write( "Erase the first two 0 terms of the third equation:\r\n" );
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document.write( " x + \"3%2F2\"y - 3z = 2\r\n" );
document.write( "        y + 0z = 2\r\n" );
document.write( "            0z = 0\r\n" );
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document.write( "The last equation is true for any value of z\r\n" );
document.write( "So we let z be arbitrary. Then z = a\r\n" );
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document.write( "Substitute in the 2nd equation:\r\n" );
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document.write( "     y + 0(a) = 2\r\n" );
document.write( "            y = 2\r\n" );
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document.write( "Substitute z = a and y = 2 in the first equation:\r\n" );
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document.write( " x + \"3%2F2\"(2) - 3(a) = 2\r\n" );
document.write( "        x + 3 - 3a = 2\r\n" );
document.write( "                 x = 3a-1\r\n" );
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document.write( "So the set of solutions is\r\n" );
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document.write( "(x, y, z) = (3a-1, 2, a)\r\n" );
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document.write( "Edwin
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