Quadratic and Fractions

The two equations can be simplified and equated. This yields:

$(3x+2)(x+1)(x-6)$ $=$ $-132$

$-132$ can be written as $11\times4\times-3$

Equating the polynomials to the factors, the only combination that is equal is when $3x+2=11$ , $x+1=4$ and $x-6=-3$

Thus, $x=\boxed{3}$

since the value of y is positive....then y=[1,2,3,4,5]........by hit and trail 3 will satisfy both the equations .

3x^2 + 5x + 2 = (x+1)(3x+2) = -132/x-6 = 132/6-x = 2 2 3*11/6-x Since x>0 and x is an integer, x only can take values 2, 3, 4 and 5. And by simple evaluation, we get x=3