document.write( "Question 1202869: Please help me solve these equations
\n" ); document.write( "1) x + y + z = 3
\n" ); document.write( "x + 2y - z = 2
\n" ); document.write( "x + yz + zx = 3
\n" ); document.write( "2) x +yz = 2
\n" ); document.write( " y + zx = 2
\n" ); document.write( " z + xy = 2 \n" ); document.write( "

Algebra.Com's Answer #838054 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "    x + yz = 2,    (1)\r\n" );
document.write( "    y + zx = 2,    (2)\r\n" );
document.write( "    z + xy = 2.    (3)\r\n" );
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document.write( "ANSWER.  There are two sets of solutions: (a) (x,y,z) = (1,1,1) and (b) (x,y,z) = (-2,-2,-2).\r\n" );
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document.write( "                                  Solution\r\n" );
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document.write( "First, it is easy to see by inspection that  x = y = z = 1  is the solution.\r\n" );
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document.write( "Next, our goal is to check if there are other solutions.\r\n" );
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document.write( "Let's assume x= 1.  Then equations (1) and (2) take the forms\r\n" );
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document.write( "    1 + yz = 2,    (1')\r\n" );
document.write( "    y + z  = 2.    (2')\r\n" );
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document.write( "From (2'), express z = 2-y and substitute it into equation (1').  You will get\r\n" );
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document.write( "    1 + y*(2-y) = 2  -->  1 + 2y - y^2 = 2  -->  y^2 - 2y + 1 = 0  -->  (y-1)^2 = 0  -->  y= 1.    (4)\r\n" );
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document.write( "Thus if we assume x= 1, then y= 1  and similarly z= 1.\r\n" );
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document.write( "In the same way, if we assume that any single variable is 1, then we get \r\n" );
document.write( "that two other variables are equal to 1, due to equations.\r\n" );
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document.write( "     So, if we are looking for other sets of solutions, \r\n" );
document.write( "       where at least one unknown is not equal to 1,\r\n" );
document.write( "    then no one of the three unknowns in these sets is 1.\r\n" );
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document.write( "    +-----------------------------------------------------+\r\n" );
document.write( "    |       OK, this step of reasoning is complete.       |\r\n" );
document.write( "    |    At this point, we start next step of reasoning.  |\r\n" );
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document.write( "From equations (1) and (2)\r\n" );
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document.write( "    x + yz = y + zx,\r\n" );
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document.write( "    (x-y) - (xz - yz) = 0,\r\n" );
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document.write( "    (x-y) - z(x-y) = 0,\r\n" );
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document.write( "    (1-z)*(x-y) = 0.       \r\n" );
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document.write( "Since we assume now that  z =/= 1,    we come to  x = y.    (5)\r\n" );
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document.write( "Similarly, from equations (1) and (3) we come to  x = z.    (6)\r\n" );
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document.write( "Similarly, from equations (2) and (3) we come to  y = z.    (7)\r\n" );
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document.write( "    +--------------------------------------------------+\r\n" );
document.write( "    |     Let before the last attack  w  denotes       |\r\n" );
document.write( "    |  any of the three equal quantities  x = y = z.   |\r\n" );
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document.write( "Then equation (1) takes the form\r\n" );
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document.write( "    w + w^2 = 2  -->  w^2 + w - 2 = 0  -->  factoring  (w-1)*(w+2) = 0  -->  w = 1  or  w = -2.\r\n" );
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document.write( "Thus we get these triples  (x,y,z) = (1,1,1)  and  (x,y,z) = (-2,-2,-2),  that we announced as the solutions at the beginning.\r\n" );
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document.write( "Easy check confirms that  (x,y,z) = (-2,-2,-2)  is the second solution set.\r\n" );
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document.write( "There is NO any other solutions.\r\n" );
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