Question

Solve the following system of inequations in $R$
$R$.
$\frac{2x-3}{4}-2\ge \frac{4x}{3}-6$,
$2(2x+3)<6(x-2)+10$

Answer 1

Let us consider the first inequation.
$\frac{2x-3}{4}-2\ge \frac{4x}{3}-6$
$\Rightarrow \frac{2x-3-8}{4}\ge \frac{4x-18}{3}$ $\Rightarrow \frac{2x-11}{4}\ge \frac{4x-18}{3}$
Multiply by 12 on both sides we get,
$\Rightarrow \left(\frac{2x-11}{4}\right)3\times 4\ge \left(\frac{4x-18}{3}\right)\times 3\times 4$
$\Rightarrow 3(2x-11)\ge 4(4x-18)$

$\Rightarrow 6x-33\ge 16x-72$
Transposing −33 to "R.H.S." and 16x to L.H.S.
$\Rightarrow 6x-16x\ge -72+33$
$\Rightarrow -10x\ge -39$
Divide by -10 on both sides we get,
$\Rightarrow \frac{-10x}{-10}\le \frac{-39}{-10}$
$[\therefore \text{Inequality sign reverses when both sides of inequation is divided by negative number.}]$
$\Rightarrow x\le \frac{39}{10}$
$\therefore (-\infty ,\frac{39}{10}]....\left(1\right)$

Now, let us consider the second inequation,
$2(2x+3)<6(x-2)+10$
$\Rightarrow 4x+6<6x-12+10$
$\Rightarrow 4x+6<6x-2$
Transposing 6 to R.H.S. and 6x to L.H.S.
$\Rightarrow 4x-6x<-2-6$
$\Rightarrow -2x<-8$
Divide by −2 on both sides we get,
$\Rightarrow \frac{-2x}{-2}>\frac{-8}{-2}$
$[\therefore \text{Inequality sign reverses when both sides of inequation is divided by negative number.}]$
$\Rightarrow x>4$
$\therefore (4,-\infty )....\left(2\right)$
From (1) and (2)we get,
$xϵ(-\infty ,\frac{39}{10})\cap (4,\infty )$
$\therefore xϵ\varphi$

Thus, there is no solution set of the given system of inequations