Find all values of $x$ such that $\frac{4x-5}{3x+5}\geq3$ Find all values of $x$ such that $\frac{4x-5}{3x+5}\geq3$.
If we consider two cases: $3x+5 > 0$ for the first and $3x + 5 < 0$ for the second.
When $3x+5 > 0$:
$\frac{4x-5}{3x+5}\geq3$.
Sloving for $x$ we find $x\leq -4$.
When $3x+5 < 0$:
$\frac{4x-5}{3x+5}\geq3$.
Sloving for $x$ we find $x\geq -4$
Additionaly through trail and error I discovered $x=-2$, $x=-3$ and $x=-4$... satisfy the original inequality.
$-2$, $-3$, $-4$ satisfy the inequality chain $-4 \leq x < -5/3$. All other values the satisify the inequality chain are also solutions.
In the case that $3x+5 > 0$, the inequality chain $-4 \geq x > -5/3$ is illogical.
Is this soultion correct? How can I improve it?

Your way to proceed also works, indeed we have

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*for $3x+5>0 \iff x>-\frac 5 3$
$$\frac{4x-5}{3x+5}\geq3 \iff 4x-5\ge 9x+15 \iff 5x \le -20 \iff x \le -4$$
that is no solution for this first case, then

*

*for $3x+5<0 \iff x<-\frac 5 3$
$$\frac{4x-5}{3x+5}\geq3 \iff 4x-5\le 9x+15 \iff 5x \ge -20 \iff x \ge -4$$
that is $x \in \left[-4,-\frac 5 3\right)$.
