document.write( "Question 1098865: The no. Of solution of (x,y,z)to the system of equations x+2y+4z=9, 4yz+2xz+xy=13,xyz=13, such that at least two of x,y,z are integers is...
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Algebra.Com's Answer #713343 by ikleyn(52794) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "The system as it is posted, HAS NO integer or rational solutions AT ALL.\r
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\n" ); document.write( "\n" ); document.write( "In order to make it VISIBLE and SOLVABLE, I changed the condition in this way: \r
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document.write( "    Solve the system\r\n" );
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document.write( "           x + 2y  + 4z = 9, \r\n" );
document.write( "        4yz + 2xz + xy = 13,\r\n" );
document.write( "              xyz      =  3.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Notice that I changed the third equation.\r
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document.write( "Introduce new variables \r\n" );
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document.write( "p = x,  q = 2y  and  r = 4z.           (1)\r\n" );
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document.write( "Then  pq = 2xy,  pr = 4xz,  and  qr = 8yz,  so  pq + pr + qr = 2xy + 4xz + 8yz = 2*(xy + 2xz + 4yz).\r\n" );
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document.write( "Therefore, the original system takes the form\r\n" );
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document.write( "p + q + r = 9,          (2)    \r\n" );
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document.write( "pq + pr + qr = 26,      (3)    (26 = 2*13)\r\n" );
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document.write( "pqr = 24.               (4)    (24 = 8*3)\r\n" );
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document.write( "Thus, according to the Vieta's theorem,  p, q and r are the roots of this polynomial equation of the degree 3:\r\n" );
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document.write( "t^3 - 9t^2 + 26t  - 24 = 0.\r\n" );
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document.write( "By applying the \"Rational root theorem\", you can find the integer roots of this equation.\r\n" );
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document.write( "If exist, they are among the integer divisors of the constant term 24, i.e. among the number set\r\n" );
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document.write( "{+/-1, +/-2, +/-3, +/-6, +/-12, +/-24}.\r\n" );
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document.write( "Or, you can plot a graph, and it will tell you/ (will show you) what the roots are:\r\n" );
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document.write( "Plot y = \r\n" );
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document.write( "and the plot says that the roots are t = 2, 3 and 4.  // You can check it manually.\r\n" );
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document.write( "Thus the solution (p,q,r) to the system  (2), (3), (4)  is ANY permutation of numbers  (2,3,4).\r\n" );
document.write( "Any of 3! = 1*2*3 = 6 possible permutations.\r\n" );
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document.write( "Now we must return from variables p, q and r to the original variables x, y and z using formulas (1).\r\n" );
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document.write( "Since the original system is EQUIVALENT to the system (2), (3), (4) due to substitution (1),\r\n" );
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document.write( "it implies that the following values (triples) of x, y and z are the solutions to the original system:\r\n" );
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document.write( "    1)  (p,q,r) = (2,3,4)  ====> x= 2,  y= 3/2,  z=  1;\r\n" );
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document.write( "    2)  (p,q,r) = (4,2,3)  ====> x= 4,  y= 1,    z= 3/4;\r\n" );
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document.write( "    3)  (p,q,r) = (3,4,2)  ====> x= 3,  y= 2,    z= 1/2;\r\n" );
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document.write( "    4)  (p,q,r) = (3,2,4)  ====> x= 3,  y= 1,    z=  1;  \r\n" );
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document.write( "    5)  (p,q,r) = (4,3,2)  ====> x= 4,  y= 3/2,  z= 1/2;\r\n" );
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document.write( "    6)  (p,q,r) = (2,4,3)  ====> x= 2,  y= 2,    z= 3/4.\r\n" );
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document.write( "You can check it manually (as I did it using Excel in my computer).\r\n" );
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document.write( "Answer.  The given system has 6 solutions listed above.\r\n" );
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