document.write( "Question 588053: Use row of operations to solve x+y+z=-5, x-y+3z=-1, 4x+y+z=-2
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Algebra.Com's Answer #374348 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
Here is one exactly like it\r
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document.write( "Swap row 1 and 2 to get a 1 in the upper left corner\r\n" );
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document.write( "Multiply Row 1 temporarily by -2, getting \r\n" );
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document.write( "\"%28matrix%281%2C+5%2C+-2%2C2%2C+-2%2C+%22%7C%22%2C+0%29%29\"\r\n" );
document.write( "and add that to row 2:\r\n" );
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document.write( "Swap rows 2 and 3:\r\n" );
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document.write( "Multiply Row 2 temporarily by -3, getting \r\n" );
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document.write( "\"%28matrix%281%2C+5%2C+0%2C+-3%2C+-6%2C+%22%7C%22%2C+-3%29%29\"\r\n" );
document.write( "and adding that to row 3:\r\n" );
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document.write( "Mutiply row 3 by \"-1%2F9\"\r\n" );
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document.write( "Convert back into a system of equations:\r\n" );
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document.write( "\"system%28x-y%2Bz=0%2C+0x%2B1y%2B2z=1%2C0x%2B0y%2B1z=0%29\"\r\n" );
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document.write( "or just\r\n" );
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document.write( "\"system%28x-y%2Bz=0%2C+y%2B2z=1%2Cz=0%29\"\r\n" );
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document.write( "Substitute \"z=0\" from the bottom equation\r\n" );
document.write( "into the middle equation \"y%2B2z=1\"\r\n" );
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document.write( "\"y%2B2%280%29=1\"\r\n" );
document.write( "\"y%2B0=1\"\r\n" );
document.write( "\"y=1\"\r\n" );
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document.write( "Substitute \"z=0\" and \"y=1\"\r\n" );
document.write( "into the top equation \"x-y%2Bz=0\"\r\n" );
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document.write( "\"x-%281%29%2B%280%29=0\"\r\n" );
document.write( "\"x-1=0\"\r\n" );
document.write( "\"x=1\"\r\n" );
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document.write( "So the solution is \"%22%28x%2Cy%2Cz%29%22=%22%281%2C1%2C0%29%22\"\r\n" );
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document.write( "Edwin
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