Question

Solve the following system of inequations in $R$
$R$.
$\frac{7x-1}{2}<-3,$
$\frac{3x+8}{5}+11<0$

Answer 1

Given system of inequations:
$\frac{7x-1}{2}<-3and\frac{3x+8}{5}+11<0$
Let us consider the first inequation.
$\frac{7x-1}{2}<-3$
Multiplying by 2 on both sides we get,
$\Rightarrow \left(\frac{7x-1}{2}\right)\times 2<-3\times 2$
$\Rightarrow 7x-1<-6$
Adding 1 on both sides we get,
$\Rightarrow 7x-1+1<-6+1$

$\Rightarrow 7x<-5$
Divide by 7 on both sides we get,
$\Rightarrow \frac{7x}{7}<\frac{-5}{7}$
$\Rightarrow x<-\frac{5}{7}$
$\therefore xϵ(-\infty ,-\frac{5}{7})\dots \left(1\right)$
Now, let us consider the second inequation.
$\frac{3x+8}{5}+11<0$
$\Rightarrow \frac{3x+8+55}{5}<0$
$\Rightarrow \frac{3x+63}{5}<0$
Multiplying by 5 on both sides we get,
$\Rightarrow \left(\frac{3x+63}{5}\right)\times 5<0\times 5$
$\Rightarrow 3x+63<0$
Subtract 63 from both sides we get,
$\Rightarrow 3x+63-63<0-63$
$\Rightarrow 3x<-63$
Divide by 3 on both sides we get,
$\Rightarrow \frac{3x}{3}<\frac{-63}{3}$
$\Rightarrow x<-21$
$\therefore xϵ(-\infty ,-21)\dots \left(2\right)$

From (1) and (2)we get,
$xϵ(-\infty ,-\frac{5}{7})\cap (-\infty ,-21)$
$\therefore xϵ-21$

Thus, there is no solution set of the given system of inequations is $(-\infty ,-21)$