Question

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

Answer 1

Given system of inequations:
$\frac{2x+1}{7x-1}>5and\frac{x+7}{x-8}>2$
Let us consider the first inequation.
$\frac{2x+1}{7x-1}>5$
Subtract 5 from both sides we get,
$\Rightarrow \frac{2x+1}{7x-1}-5>5-5$
$\Rightarrow \frac{2x+1-5(7x-1)}{7x-1}>0$
$\Rightarrow \frac{2x+1-35x+5}{7x-1}>0$
$\Rightarrow \frac{-33x+6}{7x-1}>0$
$\Rightarrow \frac{-3(11x-2)}{7x-1}>0$
$\Rightarrow \frac{11x-2}{7x-1}<0\cdots \left(i\right)$ [Divide by −3 on both sides]
$[\therefore \text{Inequality sign reverses when both sides of inequation is divided by negative number.}]$
For critical points,
$11x-2=0and7x-1=0$
$x=\frac{2}{11}\&x=\frac{1}{7}$

Now plotting points on number line, and evaluating signs of expression on L.H.S. of inequation (𝐢) for the different regions, we have

Taking the region where expression on L.H.S of inequation (i) is -ve.
Hence $x,ϵ(\frac{1}{7},\frac{2}{11}).....\left(ii\right)$
Now, let us consider the second inequation.
$\frac{x+7}{x-8}>2$
Subtract 2 from both sides we get,
$\Rightarrow \frac{x+7}{x-8}-2>0$
$\Rightarrow \frac{x+7-2x+16}{x-8}>0$
$\Rightarrow \frac{-x+23}{x-8}>0$
$\Rightarrow \frac{x-23}{x-8}<0\cdots \left(iii\right)$
$[\therefore \text{Inequality sign reverses when both sides of inequation is divided by negative number}.]$
For critical points,
$x-23=0andx-8=0$
x=23 & x=8
Now plotting points on number line, and evaluating signs of expression on L.H.S. of inequation (iii) for the different regions, we have
Taking the region where expression on L.H.S. of inequation (iii) is -ve
Hence $xϵ(8,23)......\left(iv\right)$

From (ii) and (iv) we get,
$xϵ(\frac{1}{7},\frac{2}{11})\cap (8,23)$
$\therefore xϵ\varphi$

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