Answer
$(-∞, -3) U (-2/3, 1) U (3, ∞)$
Work Step by Step
$\frac{x^2 + 2x - 3}{3x^2 - 7x - 6} > 0$
Find the zeros of the expressions in the numerator AND the denominator
$x^2 + 2x - 3 = 0$
$(x+3) (x-1) = 0$
$x = -3, 1$
$3x^2 - 7x - 6 = 0$
$(3x + 2) (x - 3) = 0$
$x = -2/3, 3$
$\frac{(x+3) (x-1)}{(3x + 2) (x - 3)} \leq 0$
Test numbers in between those zero values to determine if the function is negative or positive
$(-∞, -3)$ $\frac{(-)(-)}{(-)(-)} = (+)$
$(-3,-2/3)$ $\frac{(+)(-)}{(-)(-)} = (-)$
$(-2/3, 1)$ $\frac{(+)(-)}{(+)(-)} = (+)$
$(1, 3)$$\frac{(+)(+)}{(+)(-)} = (-)$
$(3, ∞)$ $\frac{(+)(+)}{(+)(+)} = (+)$
Thus the solution is $(-∞, -3) U (-2/3, 1) U (3, ∞)$
