Question 1208843
Find the real solutions of the equations. 


(A) 2x^(-2) - 3x^(-1) - 4 = 0


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



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


Introduce new variable 

    t = {{{y/(y-1)}}}.


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  {{{y/y-1)}}} = 7,   y = 7*(y-1),  y = 7y - 7,     7 = 7y - y,  7 = 6y,  y = 7/6.


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


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

(B) is solved.