How to solve $(x - 1)(x + 7) < 0$ I tried this:
\begin{align*}
                (x - 1)(x + 7) < 0
                &\iff x-1<0 \vee x+7<0\\
                &\iff x < 1 \vee x < -7\iff x\in(-7,1).
            \end{align*}
The thing though, is that I'm not sure if $ab<0 \implies a<0$ or $b<0$.

A product $ab < 0$ if $a$ and $b$ have opposite signs.  When $a = x - 1$ and $b = x + 7$
$$(x - 1 > 0 \wedge x + 7 < 0) \vee (x - 1 < 0 \wedge x + 7 > 0)$$
Since $x - 1 > 0 \implies x > 1$ and $x + 7 < 0 \implies x < -7$, the statement $$x - 1 > 0 \wedge x + 7 < 0$$ is never true since it is not possible for a real number to be both greater than $1$ and less than $-7$.
Since $x - 1 < 0 \implies x < 1$ and $x + 7 > 0 \implies x > -7$, the statement $$x - 1 < 0 \wedge x + 7 > 0$$ is true when $-7 < x < 1$.
Hence, the solution set is $S = (-7, 1)$.
Another way to do the problem is to perform a line analysis.  Since $(x + 7)(x - 1) = 0$ when $x = -7$ or $x = 1$ and $(x + 7)(x - 1)$ is continuous, the sign of the product can change at the points $x = -7$ or $x = 1$.  We draw a number line with the points $x = -7$ and $x = 1$ marked as zeros.  We wish to determine the sign of the product $(x + 7)(x - 1)$ in the three intervals $(-\infty, -7), (-7, 1), (1, \infty)$.  To do so, we determine the signs of $x + 7$ and $x - 1$ in each of these intervals.  The term $x + 7$ is negative when $x < -7$, zero at $x = -7$, and positive when $x > -7$.  The term $x - 1$ is negative when $x < 1$, zero at $x = 1$, and positive when $x > 1$.  The sign of $(x + 7)(x - 1)$ in each interval is found by multiplying the signs of the factors $x + 7$ and $x - 1$ in that interval.

Since we wish to determine where $(x + 7)(x - 1) < 0$, our solution set occurs where $(x + 7)(x - 1)$ is negative, which is the interval $(-7, 1)$.
