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What is the greatest possible value of x that solves the equation \((\frac {4x - 16}{3x - 4})^2+(\frac {4x - 16}{3x - 4} )=12\)  ?
 

 Mar 28, 2019
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\(u=\dfrac{4x-16}{3x-4}\\ u^2 + u -12=0\\ (u+4)(u-3) = 0\\ u=-4,3\\ \dfrac{4x-16}{3x-4}=-4 \Rightarrow x=2\\ \dfrac{4x-16}{3x-4}=3 \Rightarrow x=-\dfrac 4 5\)

 

\(2 \text{ is the greatest possible solution}\)

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 Mar 28, 2019

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