document.write( "Question 499709: How do you find x,y,and z in the following equation?\r
\n" ); document.write( "\n" ); document.write( "5x-2y+4z=0
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\n" ); document.write( "x-y+5z=0 \n" ); document.write( "
Algebra.Com's Answer #337857 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "(1)    5x - 2y + 4z =   0\r\n" );
document.write( "(2)    5x - 4y - 3z = -21\r\n" );
document.write( "(3)     x -  y + 5z =   0\r\n" );
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document.write( "Multiply (2) by -1 and add it to (1)\r\n" );
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document.write( "       5x - 2y + 4z =  0\r\n" );
document.write( "      -5x + 4y + 3z = 21\r\n" );
document.write( "     ---------------------\r\n" );
document.write( "(4)         2y + 7z = 21\r\n" );
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document.write( "Multiply (3) by -5 and add (1)\r\n" );
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document.write( "     -5x +  5y - 25z = 0\r\n" );
document.write( "      5x -  2y +  4z = 0\r\n" );
document.write( "    ----------------------\r\n" );
document.write( "(5)         3y - 21z = 0\r\n" );
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document.write( "Multiply (4) by 3 and add (5)\r\n" );
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document.write( "            6y + 21z = 63\r\n" );
document.write( "            3y - 21z =  0\r\n" );
document.write( "          ------------------\r\n" );
document.write( "            9y       = 63  \r\n" );
document.write( "                   y = 7\r\n" );
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document.write( "Substitute y = 7 in (5)\r\n" );
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document.write( "          3(7) - 21z = 0\r\n" );
document.write( "            21 - 21z = 0           \r\n" );
document.write( "                -21z = -21\r\n" );
document.write( "                   z = 1\r\n" );
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document.write( "Substitute y = 7 and z = 1 in (3)\r\n" );
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document.write( "         x - (7) + 5(1) = 0\r\n" );
document.write( "              x - 7 + 5 = 0\r\n" );
document.write( "                  x - 2 = 0\r\n" );
document.write( "                      x = 2\r\n" );
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document.write( "(x,y,z) = (2,7,1)\r\n" );
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document.write( "Edwin
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