Example question: Solve $\displaystyle \frac{3x^2+x-7}{x-1} > x+1$ algebraically.

$\displaystyle \frac{3x^2+x-7}{x-1} - (x+1) > 0 \\
\displaystyle \frac{3x^2+x-7-(x+1)(x-1)}{x-1} > 0 \\
\displaystyle \frac{3x^2+x-7-(x^2-1)}{x-1} > 0 \\
\displaystyle \frac{2x^2+x-6}{x-1} > 0 \\
\displaystyle \frac{(x+2)(2x-3)}{x-1} > 0$

$ -2 < x < 1 $ or $ x > \frac{3}{2}$.

This is arguably the most vital (non-)step. Let us start by first recalling the rule for multiplication/division involving inequalities: the sign remains if we multiply/divide by a positive number, and the sign "flips" if we do so by a negative number.

Algebraic terms (e.g. $x$, $x-1$, $x^2+2x-3$) can be positive or negative depending what $x$ is; cross multiplying means we would have made a mistake right away.

To emphasize, DO NOT cross multiply. *(Unless you are absolutely sure of what you are doing).

Since we should not cross multiply, addition/subtraction is the way to go. This explains the first step of the worked solution above. We then simplify by forcing a common denominator and combining the terms into a single algebraic fractions.

Now that we have a nice single algebraic fraction and $0$ on the other side of the inequality, we factorize everything if possible. Thankfully this can be done for this question. What happens if we cannot do so? That is where example 1.2b comes in.

The values of $x$ that makes the numerator/denominator $0$ are the points where the algebraic fraction may change sign (i.e. whether the entire expression is positive or negative). Factorization from the previous step will have made these values of $x$ clear. We can then proceed to draw a number line with those values (and only those values) marked out.

For this example, the number line is broken into 4 different regions. We then proceed to figure out whether $\displaystyle \frac{(x+2)(2x-3)}{x-1}$ is positive or negative in those regions. For example, the "$+$" sign in between $-2$ and $1$ represents that $\displaystyle \frac{(x+2)(2x-3)}{x-1}$ is positive when $-2 < x < 1$.

There are two methods to do this. We can use a logical argument: for example, if $x$ is bigger than $\frac{3}{2}$, then $(x+2), (2x-3)$ and $(x-1)$ are all positive, so the overall expression is positive. If $x$ is between $1$ and $\frac{3}{2}$, then $(x+2)$ and $(x-1)$ are still positive but $(2x-3)$ is negative so the overall expression negative. Alternatively, we can substitute an appropriate number in the region. For example, substituting $x=2$ gives us $\frac{(2+2)(4-3)}{2-1} = 4$ which is positive while substituting $x=1.1$ gives $\frac{(1.1+2)(2.2-3)}{1.2-1} = -12.8$ which is negative.

Repeat this for all 4 regions. Since we are solving $\displaystyle \frac{(x+2)(2x-3)}{x-1} > 0$, we want the regions that are positive. This gives us our final answer shown above.