Question 1209897
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(x^2y^2-1)/(2y-1) = 4x+y
x^2y^2-1 = (4x+y)(2y-1)
x^2y^2-1 = 8xy-4x+2y^2-y
x^2y^2-1-8xy+4x-2y^2+y = 0
(x^2-2)y^2+(-8x+1)y+4x-1 = 0


Compare that to ay^2+by+c = 0
a = x^2-2
b = -8x+1
c = 4x-1
The discriminant must be 0 or larger so that we end up with real number solutions for variable y.
b^2-4ac >= 0
(-8x+1)^2-4(x^2-2)(4x-1) >= 0
-16x^3+68x^2+16x-7 >= 0


Use a graphing calculator such as <a href="https://www.desmos.com/calculator">Desmos</a> or <a href="https://www.geogebra.org/calculator">GeoGebra</a> to plot out the cubic curve f(x) = -16x^3+68x^2+16x-7 


The three approximate roots are 
p = -0.43067
q = 0.22815
r = 4.45252


Between roots q and r, we have -16x^3+68x^2+16x-7 above the x axis. 
At root r is when x is maxed out, such that the discriminant is zero and y is a real number.
If x gets any larger, then the discriminant becomes negative and leads y to being a non-real complex number.



Answer: <font color=red>x = 4.45252 (approximate)</font>
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