Question

If the following inequation is true, $\frac{2x+4}{x-1}\ge 5$. The maximum possible value of $x$ is:

Answer 1

We have $\Rightarrow \frac{2x+4}{x-1}\ge 5\Rightarrow \frac{2x+4}{x-1}-5\ge 0$
$\Rightarrow \frac{2x+4-5(x-1)}{x-1}\ge 0\Rightarrow \frac{2x+4-5x+5}{x-1}\ge 0$
$\Rightarrow \frac{-3x+9}{x-1}\ge 0$ [Multiplying both sides by - 1]
$\Rightarrow \frac{3x-9}{x-1}\le 0\Rightarrow \frac{3(x-3)}{x-1}\le 0$ [Dividing both sides by 3]
$\Rightarrow \frac{x-3}{x-1}\le 0\Rightarrow 1<x\le 3$

Hence, the maximum value is $3$.