Find positive numbers $M$, $N$ such that $\forall x\in[-1,1]$

$M\leq\left|\frac{2x+7}{5-3x}\right|\leq N$

[Comments in italics]

This problem requires the following fact. For $a,b,c,d\in\mathbb{R}$, if $0\leq a< b$ and $0\leq c< d$ then we may divide the smaller by the larger and the larger by the smaller to preserve the inequality,  i.e.

$\frac{a}{d}\leq \frac{b}{c}$

Now

$\left|\frac{2x+7}{5-3x}\right|=\frac{|2x+7|}{|5-3x|}$

Now, for the upper bound, by the triangle inequality,

$|2x+7|\leq |2x|+7=2|x|+7\leq 9$

As the maximum of $|x|$ as $x\in[-1,1]$ is 1; i.e. for $x\in[-1,1], |x|\leq 1$. We also used $|xy|=|x||y|$

By the reverse triangle inequality,

$|5-3x|\geq\left||5|-|3x|\right|=\left|5-3|x|\right|$

For $x\in[-1,1]$, $5-3|x|$ is positive [if $|x|\leq 1$, then $3|x|\leq 3$ and so $0\leq 3-3|x|\leq 5-3|x|$] so

$\left|5-3|x|\right|=5-3|x|$

$\Rightarrow |5-3x|\geq 5-3|x|\geq 2$

We have already seen $3-3|x|\geq 0$; add $2$ to both sides.

So we have $0\leq|2x+7|\leq9$ and $0\leq 2\leq|5-3x|$ hence

$\frac{|2x+7|}{|5-3x|}\leq \frac{9}{2}$

Now, for the lower bound, by the reverse triangle inequality:

$|2x+7|=|2x-(-7)|\geq \left||2x|-|-7|\right|=\left|2|x|-7\right|$

For $x\in[-1,1]$, $2|x|-7$ is negative [if $|x|\leq 1$, then $2|x|\leq 2$ and so $2|x|-7\leq -5$] so

$\left|2|x|-7\right|=7-2|x|$

$\Rightarrow |2x+7|\geq 7-2|x|\geq 2$

We have already seen $2-2|x|\geq 0$; add $5$ to both sides.

Now by similar arguments to above:

$|5-3x|\leq |5|+|-3x|=5+3|x|\leq 8$

So we have $0\leq 5\leq|2x+7|$ and $0\leq|5-3x|\leq8$ hence

$\frac{5}{8}\leq \frac{|2x+7|}{|5-3x|}$

Putting these together we get $M=5/8$ and $N=9/5$:

$\frac{5}{8}\leq\left|\frac{2x+7}{5-3x}\right|\leq \frac{9}{5}$