
Solving inequalities
Hi! Can anyone help? I need to solve the following inequalitiy
$\displaystyle \frac{{3x}^{2 }3x5 }{{x}^{2 }2x3 } >3$
I rearrange the inequalitiy first into
$\displaystyle \frac{{3x}^{2 }3x5 }{{x}^{2}2x3 }3>0 $
I the need to use the mulitiplication rule but am not sure what to multiply by...

Write $\displaystyle \frac{3x^23x5}{x^22x3}3 =\frac{3x^23x53(x^22x3)}{x^22x3}$ and factorize the denominator.

note that 3x^2  3x  5 = 3(x^2  2x  3) + 3x + 4.
so (3x^2  3x  5)/(x^2  2x  3) = 3 + (3x + 4)/(x^2  2x  3).
clearly, this will be > 3 iff (3x + 4)/(x^2  2x  3) > 0.
now (3x + 4)/(x^2  2x  3) = (3x + 4)/[(x  3)(x + 1)].
for this to be positive, the numerator and the denominator must both have the same sign.
suppose 3x + 4 > 0, that is x > 4/3.
if 4/3 < x < 1, then x+1 is negative, and x3 is also negative, so the denominator and numerator are both positive.
if 1 < x < 3, then x+1 is positive, and x3 is negative, so the denominator is negative, so these values won't work.
if 3 < x, then x+1 is positive, and x3 is positive, so the denominator and the numerator are both positive.
now, suppose x < 4/3. then x+1 and x3 are always negative, so the denominator is positive, so these values won't work.
so the only values of x for which (3x^2  3x  5)/(x^2  2x  3) > 3 are in the intervals (4/3,1), and (3,∞).

also this can be derived to $\displaystyle (x3)(x+1)(3x+4)<0$
after test points you get
$\displaystyle x>3$ and $\displaystyle \frac{4}{3 } < x < 1$