Question

Solve the following system of inequalities

$\frac{(2x+1)}{(7x-1)}>5$, $\frac{(x+7)}{(x-8)}>2$

Answer 1

Given : $\frac{(2x+1)}{(7x-1)}>5$

$\Rightarrow \frac{2x+1}{7x-1}-5>0$

$\Rightarrow \frac{2x+1-5(7x-1)}{7x-1}>0$

$\Rightarrow \frac{6-33x}{7x-1}>0$

$\Rightarrow \frac{x-\frac{2}{11}}{x-\frac{1}{7}}<0$

Using the number line

$\therefore x\epsilon (\frac{1}{7},\frac{2}{11})$ …(i)

$\because \frac{x+7}{x-8}>2$ [ Given ]

$\Rightarrow \frac{x+7}{x-8}-2>0$

$\Rightarrow \frac{x+7-2x+16}{x-8}>0$

$\Rightarrow \frac{23-x}{x-8}>0$

$\Rightarrow \frac{x-23}{x-8}<0$

Using the number line,

$\therefore x\epsilon (8,23)$ …(ii)

Taking intersection of (i) & (ii)

There is no common value of $x$.

Hence, there is no solution of given system of inequalities.