+321 Given \(3x+\dfrac{2}{x}-4=2(x+9)-\left(7-\dfrac{1}{x}\right)\) what is \(\dfrac{x^2+1}{x}?\)
+592 First, simplify the equation:
\(3x + \frac{2}{x} - 4 = 2(x + 9) - (7 - \frac{1}{x})\)
\(3x + \frac{2}{x} - 4 = 2x + 18 - 7 + \frac{1}{x}\)
\(x - 15 = -\frac{1}{x}\)
\(x^2 - 15x = -1\)
\(x^2 + 1 = 15x\)
Now plug it into \(\frac{x^2 + 1}{x}\):
\(= \frac{(15x)}{x}\)
\(= 15\)
3x + 2/x - 4 = 2(x + 9) - (7 - 1/x) ==> 3x + 2/x - 4 = 2x + 18 - 7 + 1/x ==> x + 1/x = 15 ==> x^2 + 1 = 15x. Therefore, (x^2 + 1) / x = 15x / x = 15 
