Question

Solve the following inequalities. $\frac{b+3}{5-2b}\le 4$ where $b\le \frac{17}{a}$ then find $a$.

Answer 1

Consider the given inequality $\frac{b+3}{5-2b}\le 4$ and multiply $(5-2b)$ to both sides of the inequality as shown below:

$\frac{b+3}{5-2b}\times (5-2b)\le 4(5-2b)$
$b+3\le 20-8b$

Let us now add $8b$ to both sides and subtract $3$ from both sides as follows:

$b+3+8b-3\le 20-8b+8b-3\Rightarrow b+8b+3-3\le -8b+8b+20-3\Rightarrow 9b\le 17$

Now, we divide both sides by $9$:

$\frac{9b}{9}\le \frac{17}{9}\Rightarrow b\le \frac{17}{9}$

Hence, $b\le \frac{17}{9}$.