SOLUTION: Find the real solutions of the equations. A. 2x^(-2) - 3x^(-1) - 4 = 0 B. [y/(y - 1)]^2 = 6[y/(y - 1)] + 7

(B)  [y/(y - 1)]^2 = 6[y/(y - 1)] + 7.


Introduce new variable 

    t = .


Then equation (1) takes the form

    t^2 = 6t + 7.


Write it in the standard form quadratic equation 

    t^2 - 6t - 7 = 0.


Factor left side

    (t-7)*(t+1) = 0.


The roots are  t= 7  and  t= -1.


Consider two cases.


    1)  t= 7.   It means   = 7,   y = 7*(y-1),  y = 7y - 7,     7 = 7y - y,  7 = 6y,  y = 7/6.


    2)  t= -1.  It means   = -1,  y = (-1)*(y-1),  y = -y + 1,  y + y  = 1,  2y = 1,  y = 1/2.


ANSWER.  Equation (B)  has two solutions:  y= 1/2  and  y= 7/6.