Solving basic inequality I have the following steps to a solution
$1.$ $\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$
$2.$ $10 ≥ 15 - 15x+ (-12)x + 27x ≥ 5$
$3.$ $x ≥ \frac 5{27}$
I am confused about the transition between step $1$ and $2$.
Wouldn't multiplying $(1-x)$ to both sides of the inequality get me
$10 ≥ 15(1 - x) + (-12x)$
$10 ≥ 15 - 27x$
What am I doing wrong instead of getting the result in step $2$?
 A: Let's start here:
$$\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$$
You are mostly right.  Multiplying by $1-x$ results in
$$10 \ge 15(1-x) -12x, \quad \text{ whenever } x<1,\quad(1)$$ and
$$10 \le 15(1-x) -12x, \quad \text{ whenever } x>1.\quad(2)$$
Multiplying an inequality by a positive number maintains the inequality, multiplying by a negative number switches the direction of the inequality.
Both parts of (2) cannot be true, so we are left with:
$$10 \ge 15-27x \texttt{ AND } x<1$$
$$\frac{5}{27} \le x <1.$$
A: Most probably, in step $2$, it is
$10+27x\geq15-15x+(-12)x+27x \implies27x\geq5$.
If these steps are mentioned in a book, then it's more likely that there's an error in the print. Otherwise, you did the whole thing right.
Also, the above inequality holds if $x<1$.
A: You should avoid multiplying terms like $1-x$ unless you are sure about its sign. Instead, combine the fractions:
$$
0 \leqslant \frac{10}{1-x}-15 - \frac{-12x}{1-x} = \frac{10+12x}{1-x} + \frac{-15+15x}{1-x}=\frac{27x-5}{1-x} \\
\iff (27x-5)(x-1)\leqslant 0 \text{ and } x \neq 1.
$$
Therefore
$$
\frac{5}{27} \leqslant x < 1. \blacksquare
$$
