What is the greatest possible value of x that solves the equation \((\frac {4x - 16}{3x - 4})^2+(\frac {4x - 16}{3x - 4} )=12\) ?
\(u=\dfrac{4x-16}{3x-4}\\ u^2 + u -12=0\\ (u+4)(u-3) = 0\\ u=-4,3\\ \dfrac{4x-16}{3x-4}=-4 \Rightarrow x=2\\ \dfrac{4x-16}{3x-4}=3 \Rightarrow x=-\dfrac 4 5\)
\(2 \text{ is the greatest possible solution}\)
+239 
