How do you solve -7/(x+5)<=-8/(x+6)?

Mar 18, 2018

$\left\{x < - 6\right\} \cup \left\{- 5 < x \le 2\right\}$

Explanation:

$- \frac{7}{x + 5} \le - \frac{8}{x + 6}$

Let's multiply both sides by $- 1$. Since we are multiplying/dividing by a negative value, we must flip the direction of the inequality:

$\frac{7}{x + 5} \ge \frac{8}{x + 6}$

Multiply both sides by $\left(x + 5\right)$:

$7 \ge \frac{8 \left(x + 5\right)}{x + 6}$

Multiply both sides by $\left(x + 6\right)$:

$7 \left(x + 6\right) \ge 8 \left(x + 5\right)$

Distribute $7$ into $x + 6$ and distribute $8$ into $x + 5$:

$7 x + 42 \ge 8 x + 40$

Subtract $7 x$ from both sides:

$42 \ge x + 40$

Subtract $40$ from both sides:

$2 \ge x$

Hence:

$\implies \textcolor{b l u e}{x \le 2}$

Now we have to assess the actual inequality for values that $x$ cannot be due to the rational terms being undefined.

Looking at the denominator $x + 5$, we see $x \setminus \ne - 5$.

Looking at the denominator $x + 6$, we see $x \setminus \ne - 6$.

So the final result is:

$\textcolor{g r e e n}{\left\{x < - 6\right\} \cup \left\{- 5 < x \le 2\right\}}$

=======================EDIT=======================
Should mention what Douglas mentioned in the comments to make this result clearer.

After we have determined that $x \setminus \ne - 5$ and $x \setminus \ne - 6$, you should investigate the regions between this "critical" points to determine if that region is included in the final solution. In this case, the example Douglas gives of $- 5.5$ in the comments suffices to show that the region $- 6 < x < - 5$ is not a part of the solution.