document.write( "Question 1210378: Find all ordered pairs x, y of real numbers such that x+y=10 and x^3+y^3=300.
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Algebra.Com's Answer #852139 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Your starting equations are\r\n" );
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document.write( "    x   + y   =  10,      (1)\r\n" );
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document.write( "    x^3 + y^3 = 300.      (2)\r\n" );
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document.write( "In equation (2), use the decomposition of the sum of cubes in the left side\r\n" );
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document.write( "    x^3 + y^3 = (x+y)*(x^2 - xy + y^2).\r\n" );
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document.write( "In this decomposition, replace the factor  (x+y)  by 10, based on equation (1).\r\n" );
document.write( "You will get then\r\n" );
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document.write( "    x^2 - xy + y^3 = 30.   (3)    (after dividing both sides by 10)\r\n" );
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document.write( "So, now you have equivalent system of equations\r\n" );
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document.write( "    x + y = 10,            (4)\r\n" );
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document.write( "    x^2 - xy + y^2 = 30,   (5)\r\n" );
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document.write( "but the degree is lowered from 3 to 2, which is good.\r\n" );
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document.write( "Now, from equation (4) express  y = 10-x  and substitute it into equation (5).  You will get\r\n" );
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document.write( "    x^2 - x(10-x) + (10-x)^2 = 30,\r\n" );
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document.write( "    x^2 - 10x + x^2 + 100 - 20x + x^2 = 30,\r\n" );
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document.write( "    3x^2 - 30x + 70 = 0.\r\n" );
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document.write( "Use the quadratic formula\r\n" );
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document.write( "     =  =  = .\r\n" );
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document.write( "Two values for x are   = 6.290994449 (approximately)  and   = 3.709005551  (approximately).\r\n" );
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document.write( "The ordered pairs are  (x,y) = ( ,  ) = ( , )\r\n" );
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document.write( "                  and  (x,y) = ( ,  ) = ( , ).\r\n" );
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document.write( "You may check that  6.290994449^3 + 3.709005551^3 = 300.0000000205... , so the approximate solution is very good.\r\n" );
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document.write( "You also may check that exact solutions (x,y) satisfy equations (1) and (2) precisely.\r\n" );
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