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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 $2$?

Book Of Flames
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dev_el
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    If step 2 is taken verbatim from a book or other source, it can only be because of an error in the source. The "$27x\ge5$" should not be added onto the "$10\ge15-15x+(-12)x$." It is instead its own separate step (from which the final step, $x\ge{5\over27}$, follows). In short, it looks like a weird typesetting error. – Barry Cipra Nov 05 '20 at 03:09

3 Answers3

4

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:

$$10 \ge 15-27x \texttt{ AND } x<1$$

$$\frac{5}{27} \le x <1.$$

mjw
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1

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 $x<1$.

jimjim
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DeBARtha
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1

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 $$ \frac{5}{27} \leqslant x < 1. \blacksquare $$

Neat Math
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