document.write( "Question 1208843: Find the real solutions of the equations. \r
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\n" ); document.write( "\n" ); document.write( "A. 2x^(-2) - 3x^(-1) - 4 = 0\r
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Algebra.Com's Answer #847326 by ikleyn(52879) $\"\"$ $\"About$

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Find the real solutions of the equations. \r
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\n" ); document.write( "\n" ); document.write( "(B) [y/(y - 1)]^2 = 6[y/(y - 1)] + 7
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\n" ); document.write( "\n" ); document.write( " I will solve equation (B).\r
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document.write( "(B)  [y/(y - 1)]^2 = 6[y/(y - 1)] + 7.\r\n" );
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document.write( "Introduce new variable \r\n" );
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document.write( "    t = .\r\n" );
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document.write( "Then equation (1) takes the form\r\n" );
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document.write( "    t^2 = 6t + 7.\r\n" );
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document.write( "Write it in the standard form quadratic equation \r\n" );
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document.write( "    t^2 - 6t - 7 = 0.\r\n" );
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document.write( "Factor left side\r\n" );
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document.write( "    (t-7)*(t+1) = 0.\r\n" );
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document.write( "The roots are  t= 7  and  t= -1.\r\n" );
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document.write( "Consider two cases.\r\n" );
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document.write( "    1)  t= 7.   It means   = 7,   y = 7*(y-1),  y = 7y - 7,     7 = 7y - y,  7 = 6y,  y = 7/6.\r\n" );
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document.write( "    2)  t= -1.  It means   = -1,  y = (-1)*(y-1),  y = -y + 1,  y + y  = 1,  2y = 1,  y = 1/2.\r\n" );
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document.write( "ANSWER.  Equation (B)  has two solutions:  y= 1/2  and  y= 7/6.\r\n" );
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