Question

Solve the following inequalities. $\frac{-x+8}{x-2}\ge 5$ where $x\le p$ then find $p$

Answer 1

Consider the given inequality $\frac{-x+8}{x-2}\ge 5$ and multiply $(x-2)$ to both sides of the inequality as shown below:

$\frac{-x+8}{x-2}\times (x-2)\ge 5(x-2)$

$\Rightarrow -x+8\ge 5x-10$

Let us now add $x$ and $10$ to both sides as follows:

$-x+8+x+10\ge 5x-10+x+10\Rightarrow -x+x+8+10\ge 5x+x-10+10\Rightarrow 18\ge 6x\Rightarrow 6x\le 18$

Now, we divide both sides by $6$:

$\frac{6x}{6}\le \frac{18}{6}\Rightarrow x\le 3$

Hence, $x\le 3$.