Algebra.Com's Answer #816907 by ikleyn(52781)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "  =  (1)\r\n" );
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document.write( "  = 1. (2)\r\n" );
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document.write( "Introduce new variables u =  , v =  . with these variables, the equations take the form\r\n" );
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document.write( " 3u - 4v = (3)\r\n" );
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document.write( " 2u - 5v = 1. (4)\r\n" );
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document.write( "To solve it use the Elimination method.\r\n" );
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document.write( " 6u - 8v =  (5)\r\n" );
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document.write( " 6u - 15v = 3 (6)\r\n" );
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document.write( " ------------------------------- Subtract eq(5) from eq(6)\r\n" );
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document.write( " -7v = 2 =  \r\n" );
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document.write( " v =  .\r\n" );
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document.write( "Substitute v = -1/3 into equation (6) to get \r\n" );
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document.write( " 6u = 3 - 5 = -2,\r\n" );
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document.write( " u =  = .\r\n" );
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document.write( "ANSWER. x =  =  = -3. y =  = = -3.\r\n" );
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document.write( "Solved.\r
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document.write( "Introducing new variables is one of standard methods solving such equations.\r
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document.write( "To see many other similar solved problems, look into the lesson\r
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document.write( " - Solving systems of non-linear equations by reducing to linear ones \r
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document.write( "in this site.\r
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