Question

Solve the following system of inequations in $R.$

$2x-7<5-x,11-5x\le 1$

Answer 1

Given system of inequations:

$2x-7<5-x,and11-5x\le 1$

Let us consider the first inequation.

$2x-7>5-x$

Transposing $-7$ to R.H.S. and $-x$ to L.H.S.

$\Rightarrow 2x+x>5+7$

$\Rightarrow 3x>12$

Divide by 3 on both sides we get,

$\Rightarrow \frac{3x}{3}>\frac{12}{3}$

$\Rightarrow x>4$

$\therefore x\in (4,\infty )\dots \left(1\right)$

Now, let us consider the second inequation.

$11-5x\le 1$

Transposing 11 to R.H.S.

$\Rightarrow -5x\le 1-11$

$\Rightarrow -5x\le -10$

Divide by $-5$ on both sides we get,

$\Rightarrow \frac{-5x}{-5}\ge \frac{-10}{-5}$

$[\because \begin{array}{c}\text{Inequality sign reverses when both sides of}\\ \text{inequation is divided by negative number.}\end{array}]$

$\Rightarrow x\ge 2$

$\therefore x\in [2,\infty )\dots \left(2\right)$

From (1) and (2) we get,

$x\in (4,\infty )\cap [2,\infty )$


Thus, the solution set of the given system of inequations is $(4,\infty ).$