Rational Inequalities Error 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 $x>2/3$.

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.
