Question

Solve the system of inequalities: $\frac{x+7}{x-8}>2,\frac{2x+1}{7x-1}>5$.

• A. $(-16/7,5/7)$
• B. $(-7/8,2/7)$
• C. $(2,5)$
• D. No solution

Answer 1

The correct option is D No solution

$\frac{x+7}{x-8}>2$
$\frac{x+7}{x-8}-2>0$
$\frac{x+7-2x+16}{x-8}>0$
$\frac{-x+23}{x-8}>0$
$\frac{x-23}{x-8}<0$ $(a<b=>-a>-b)$
Values of x satisfying this inequality are:
$x\in (8,23)$
Now:
$\frac{2x+1}{7x-1}>5$
$\frac{2x+1}{7x-1}-5>0$
$\frac{2x+1-35x+5}{7x-1}>0$
$\frac{-33x+6}{7x-1}>0$
$\frac{11x-2}{7x-1}<0$ $(a<b=>-a>-b)$
$\frac{x-\frac{2}{11}}{x-\frac{1}{7}}<0$
Values of x satisfying this inequality are:
$x\in (\frac{1}{7},\frac{2}{11})$
We can see from the solution sets that no value of x satisfy both the inequality.