Answer
$(-\infty,-3) \cup\left( -\frac{2}{3}, 1\right) \cup (3,\infty)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{x^2+2x-3}{3x^2-7x-6} \gt 0$
$f(x)=\displaystyle \frac{(x+3)(x-1)}{(x-3)(3x+2)} \gt 0$
2. The cut points are:
$\displaystyle \displaystyle \frac{(x+3)(x-1)}{(x-3)(3x+2)} = 0$
$x=-3$ or $x=-\frac{2}{3}$ or $x=1$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \gt 0 ? \\
& & \displaystyle \displaystyle \frac{(a+3)(a-1)}{(a-3)(3a+2)} & \\
(-\infty,-3) & -5 & \frac{(-)(-)}{(-)(-)}=(+) & T\\
(-3, -\frac{2}{3}) & -2 & \frac{(+)(-)}{(-)(-)}=(-) & F\\
( -\frac{2}{3}, 1) & 0& \frac{(+)(-)}{(-)(+)}=(+) & T\\
(1, 3) & 2 & \frac{(+)(+)}{(-)(+)}=(-) & F\\
(3,\infty) & 10 & \frac{(+)(+)}{(+)(+)}=(+) & T\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty,-3) \cup\left( -\frac{2}{3}, 1\right) \cup (3,\infty)$
