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I'm slowly working my way through Serge Lang's Basic Mathematics and I'm having difficulty with this solution to a basic inequality:

$\qquad(1)$

$${-2x+5\over x+3}<1 $$

I started by making two assumptions about the denominator, which must be either greater than or less than zero. Hopefully one of these assumptions would directly contradict the simplified inequality (1) and the other would be correct, thereby defining the endpoints of the inequality. For example:

$(i)$ Assume $x+3 < 0$

$$x+3<0 \Rightarrow x<-3$$

Multiplying RHS of (1) by $(x+3)$ would reverse sign

$$-2x+5 > x+3 $$

$$-3x>-2$$

Dividing by $-3$ would reverse sign once more

$$x<2/3$$

(ii) Assume $x+3>0$

$$x+3>0 \Rightarrow x>-3$$

Multiply by RHS

$$-2x+5 < x+3 $$

$$-3x<-2$$

Dividing by $(-3)$ would reverse sign $$x>2/3$$

I'm thinking I've made some algebraic or logical mistake since the answer cannot be $x\in\mathbb{Z}$ and $x\neq \pm3$ since the original inequality would breakdown at $x=-1,-2,etc$ which means (i) is false, and it must be $x>2/3$.

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You went from $$-2x+5 > x+3$$ to $$-3x > -3$$ which is an arithmetic error.


Let's suppose for the moment that $3-5 = -3$ and examine what you did. You showed the following two facts:

  • if $x < -3$ and $\frac{-2x+5}{x+3} < 1$ then $x<1$
  • if $x > -3$ and $\frac{-2x+5}{x+3} < 1$ then $x > 1$.

So you have shown that if $\frac{-2x+5}{x+3} < 1$ then $x < 1$ or $x > 1$. (Conditional on the false fact that $3-5 = -3$.)

Now, having found some candidate solutions, you need to check that they all actually are solutions: you need to check that $x<1$ or $x>1$ implies $\frac{-2x+5}{x+3} < 1$. You can do this easily by noting that your original reasoning all reverses.

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  • $\begingroup$ I have since edited my post for the error, thanks. I must say however that the book's method was to compare the assumption (say from (i), $x<-3$, to it's result, $x<2/3$, and that one of the results would directly contradict the assumption, and the other would be correct, thereby defining the endpoints to the set. It's odd that he wouldn't include any other examples, but I suspect they may come later in the graphing sections. Nevertheless the correct answer is correct, regardless of method! $\endgroup$ – Watson Feb 18 '17 at 17:47
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In addition to @PatrickStevens' answer, I feel an easier way to tackle the inequality would be: $$\frac {5-2x}{x+3} < 1\Leftrightarrow \frac {5-2x}{x+3}-1 < 0$$ $$\Leftrightarrow \frac{2-3x}{x+3} < 0$$ $$\Leftrightarrow \frac {x-\frac {2}{3}}{x+3} > 0$$

It will now be easy to calculate the solution set. Hope it helps.

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