Question

Solve the following system of inequations in $R.$

$2x+5\le 0,x-3\le 0$

Answer 1

Given system of inequations:

$2x+5\le 0,x-3\le 0$

Let us consider the first inequation.

$2x+5\le 0$

Subtracting 5 from both sides we get,

$\Rightarrow 2x\le -5$

Divide by 2 on both sides we get,

$\Rightarrow \frac{2x}{2}\le \frac{-5}{2}$

$\Rightarrow x\le -\frac{5}{2}$

$\therefore x\in (-\infty ,-\frac{5}{2}]\dots \left(1\right)$

Now, let us consider the second inequation.

$x-3\le 0$

$\Rightarrow x\le 3\left[\text{on transposing -3 to R.H.S.}\right]$

$\therefore x\in (-\infty ,3]\dots \left(2\right)$

$\therefore x\in (-\infty ,-\frac{5}{2}]\cap (-\infty ,3]$

$\therefore x\in (-\infty ,-\frac{5}{2}]$


Thus, the solution set of the given system of inequations is $(-\infty ,-\frac{5}{2}].$