How to efficiently solve an inequality where the variable occurs in the denominator? For example, this inequality:
$$\frac{2x(x-4)}{x-1} \le 7$$
I can solve this by finding the 'critical values' (or in other words, two values which $x$ can/or cannot equal), then putting them on a number line and testing values on either side of these two numbers. 
However, I don't understand how you would solve this inequality by using the "multiply by the square of the denominator, then sketch a parabola and use the x-intercepts to solve" method.
Nor do I understand how to solve this following the 'two cases' rule where if $\frac ab \le 0$ then either 1. $a<0, b>0$ or 2. $a>0, b<0$.
Feel free to tell me any specific 'prerequisite knowledge' I should have or other concepts that I should be confident with to solve inequalities with a variable in the denominator.
By the way, please explain every step (explain why you would do it) included in your method because I'm terrible at maths. I can't wrap my head around the concept of inequalities. 
 A: Divide into two cases:


*

*$x-1>0$: Multiply each side by $x-1$, and solve $2x(x-4)\leq7(x-1)$

*$x-1<0$: Multiply each side by $x-1$, and solve $2x(x-4)\geq7(x-1)$


To summarize this, you will have to compute the range of valid values for $x$ as follows:
$$([x-1>0]\cap[2x(x-4)\leq7(x-1)])\cup([x-1<0]\cap[2x(x-4)\geq7(x-1)])$$

Let's solve the left half:


*

*$[x-1>0]\cap[2x(x-4)\leq7(x-1)]\implies$

*$[x-1>0]\cap[2x^2-8x-7x+7\leq0]\implies$

*$[x-1>0]\cap[2x^2-15x+7\leq0]\implies$

*$[x-1>0]\cap([x\leq\frac{15+\sqrt{15^2-4\cdot2\cdot7}}{2\cdot2}]\cap[x\geq\frac{15-\sqrt{15^2-4\cdot2\cdot7}}{2\cdot2}])\implies$

*$[x-1>0]\cap([x\leq7]\cap[x\geq\frac12])\implies$

*$[1<x\leq7]$

Let's solve the right half:


*

*$[x-1<0]\cap[2x(x-4)\geq7(x-1)]\implies$

*$[x-1<0]\cap[2x^2-8x-7x+7\geq0]\implies$

*$[x-1<0]\cap[2x^2-15x+7\geq0]\implies$

*$[x-1<0]\cap([x\geq\frac{15+\sqrt{15^2-4\cdot2\cdot7}}{2\cdot2}]\cup[x\leq\frac{15-\sqrt{15^2-4\cdot2\cdot7}}{2\cdot2}])\implies$

*$[x-1<0]\cap([x\geq7]\cup[x\leq\frac12])\implies$

*$[x\leq\frac12]$

So the range of valid values is $[x\leq\frac12]\cup[1<x\leq7]$.
A: You can multiply both sides by the positive number $(x-1)^2$, resulting in the inequation:
$$2x^3-17x^2+22x-7<0.$$
Then graph the cubic curve (it's not a parabola) $y=2x^3-17x^2+22x-7$. The solutions are the abscissae of the points of the curve that are below the $x$-axis.
A: Quadratic inequalities
To find a solution to
$$ax^2+bx+c \le 0$$
(or $\ge 0$) you need to determine the roots by solving $ax^2 + bx + c = 0$ for $x$. Now if there are two roots $x_1$ and $x_2$ with $x_1<x_2$, the real line is divided into three intervals,
$$\mathbb R = (-\infty, x_1] \cup [x_1, x_2] \cup [x_2,\infty)$$
Now look at the leading coefficient $a$. The sign of $a$ determines the sign of the quadratic in the outer intervals. The inner interval has reversed sign. If there is only one (or no) root, the inner interval vanishes and the sign is determined by $a$ on all of $\mathbb R$. A table:
$$
\begin{array}{c|cc}
\text{ineq.}&a>0&a<0\\
\hline
ax^2+bx+c \ge 0&(-\infty, x_1] \cup [x_2, \infty)&[x_1, x_2]\\
ax^2+bx+c\le 0 &[x_1,x_2]&(-\infty, x_1]\cup[x_2,\infty)
\end{array}
$$
The specific inequality
Generally, this will involve a case differentiation. I'll walk through your example to illustrate. Note that $a < b \Leftrightarrow \cases{ac<bc & if $c>0$\\ ac>bc & if $c<0$}$.
$$\frac{2x(x-4)}{x-1} \le 7$$
Case 1: $x < 1$
$$2x(x-4) \ge 7(x-1) \Leftrightarrow 2x^2 - 15x - 1 \ge 0$$
$\Rightarrow x\in(-\infty, \frac12] \cup [7, \infty)$. Since we assumed $x< 1$ we obtain the solution
$$((-\infty, \frac12]\cup[7,\infty)) \cap (-\infty, 1) = (-\infty, 1)$$.
Case 2: $x > 1$
$$2x(x-4) \le 7(x-1) \Leftrightarrow 2x^2 - 15x - 1 \le 0$$
The solution to the quadratic gives us $[\frac12, 7]$, so in total we get
$$[\frac12 , 7] \cap (1, \infty) = (1,7]$$
The complete solution set is now given by the union of all cases:
$$(-\infty,1) \cup (1,7] = (-\infty, 7] \setminus \{1\}$$
