document.write( "Question 1087856: Hi,\r
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\n" ); document.write( "\n" ); document.write( "GIven that y is always positive and z is always negative, solve 2xyz + yz > 3xyz for x\r
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Algebra.Com's Answer #702143 by ikleyn(52778) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "You are given \r\n" );
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document.write( "2xyz + yz > 3xyz,    (1)\r\n" );
document.write( "y > 0,               (2)\r\n" );
document.write( "z < 0.               (3)\r\n" );
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document.write( "Add -3xyz to both sides of inequality (1). You will get an equivalent inequality\r\n" );
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document.write( "-xyz + yz > 0.\r\n" );
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document.write( "Factor out yz in the last inequality. You will get an equivalent inequality\r\n" );
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document.write( "yz*(-x+1) > 0.\r\n" );
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document.write( "Divide both sides by yz.  Since yz is NEGATIVE, do not forget to replace the inequality sign by the opposite.\r\n" );
document.write( "You will get an equivalent inequality\r\n" );
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document.write( "-x + 1 < 0.\r\n" );
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document.write( "Add x to both sides.  You will get an equivalent inequality\r\n" );
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document.write( "x > 1.\r\n" );
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document.write( "It is your \r\n" );
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\n" ); document.write( "\n" ); document.write( "Answer. x > 1.\r
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\n" ); document.write( "The other's tutor writing is incorrect. For your safety, do not consider it as a valid sample.\r
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\n" ); document.write( "\n" ); document.write( "It is invalid, actually (and unfortunately). Simply ignore it.\r
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