document.write( "Question 976106: Please solve and show work. Use the Gauss-Jordan method to solve the system of equations.\r
\n" ); document.write( "\n" ); document.write( "x+y+z= -1
\n" ); document.write( "x - y + 3z = -7
\n" ); document.write( "4x+y+z= -7\r
\n" ); document.write( "\n" ); document.write( "x =
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\n" ); document.write( "z =\r
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Algebra.Com's Answer #597803 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!
Instead of doing your problem for you, I'll do one just like it and
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document.write( "Write that as a matrix by dropping the letters\r\n" );
document.write( "and putting vertical line instead of equal signs:\r\n" );
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document.write( "The idea is to get three zeros in the three positions\r\n" );
document.write( "in the lower left corner of the matrix, where the elements\r\n" );
document.write( "I've colored red are:\r\n" );
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document.write( "To get a 0 where the red 2 on the left of the middle row is,\r\n" );
document.write( "multiply R1 by -2 and add it to 1 times R2, and put it in place \r\n" );
document.write( "of the present R2.  That's written as\r\n" );
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document.write( "-2R1+1R2->R2\r\n" );
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document.write( "To make it easy, write the multipliers to the left of the two\r\n" );
document.write( "rows you're working with; that is, put a -2 by R1 and a 1 by R2\r\n" );
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document.write( "We are going to change only R2.  Although R1 gets multiplied\r\n" );
document.write( "by -2 we are going to just do that mentally and add it to R2, but\r\n" );
document.write( "not really change R1.\r\n" );
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document.write( "-----\r\n" );
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document.write( "To get a 0 where the lower left red 4 is, multiply R1\r\n" );
document.write( "by -4 and add it to 1 times R3.  That's written as\r\n" );
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document.write( "-4R1+1R3->R3\r\n" );
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document.write( "Write the multipliers to the left of the two rows you're \r\n" );
document.write( "working with; that is, put a -4 by R1 and a 1 by R3\r\n" );
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document.write( "We are going to change only R3. \r\n" );
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document.write( "---------------\r\n" );
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document.write( "To get a 0 where the red -2 is, multiply R2\r\n" );
document.write( "by -2 and add it to 3 times R3.  That's written as\r\n" );
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document.write( "-2R2+3R3->R3\r\n" );
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document.write( "Write the multipliers to the left of the two\r\n" );
document.write( "rows you're working with; that is, put a -2 by R2 and a 3 by R3\r\n" );
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document.write( "We are going to change only R3. \r\n" );
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document.write( "Now that we have 0's in the three positions in the\r\n" );
document.write( "lower left corner of the matrix, we change the matrix\r\n" );
document.write( "back to equations:\r\n" );
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document.write( "Solve the third equation for z:\r\n" );
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document.write( "Substitute 5 for z in the middle equation:\r\n" );
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document.write( "Substitute 5 for z and 3 for y in the top equation:\r\n" );
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document.write( "So the solution is \r\n" );
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document.write( "Edwin

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