Question

Solve the following system of inequations in $R$
$5x-7<3(x+3)$
$1-\frac{3x}{2}\ge x-4$

Answer 1

Given system of inequations:
$5x-7<3(x+3)and1-\frac{3x}{2}\ge x-4$
Let us consider the first inequation.
$5x-7<3(x+3)$
$\Rightarrow 5x-7<3x+9$
Transposing −7 to "R.H.S." and 3x to L.H.S.
$\Rightarrow 5x-3x<9+7$
$\Rightarrow 2x<16$
$\Rightarrow \frac{2x}{2}<\frac{16}{2}$
$\Rightarrow x<8$
$\therefore xϵ(-\infty ,8)\dots \left(1\right)$

Now, let us consider the second inequation.
$1-\frac{3x}{2}\ge x-4$
$\Rightarrow \frac{2-3x}{2}\ge x-4$
Multiplying 2 on both sides we get,
$\Rightarrow 2-3x\ge 2x-8$
Transposing 2 to R.H.S. and 2x to L.H.S.
$\Rightarrow -3x-2x\ge -8-2$
$\Rightarrow -5x\ge -10$
$\Rightarrow \frac{-5x}{-5}\le \frac{-10}{-5}$
$[\therefore \text{Inequality sign reverses when both sides of inequation is divided by negative number.}]$ $\Rightarrow x\le 2$
$\therefore xϵ(-\infty ,2]\dots \left(2\right)$
From (1) and (2)we get,
$xϵ(-\infty ,8)\cap (-\infty ,2]$
$\therefore xϵ(-\infty ,2]$

Thus, the solution set of the given system of inequations is $(-\infty ,2)$