Question

Solve the inequality. Write the solution set in interval notation.
$|5-4x|>8$

• A. $\left(\frac{3}{4},-\frac{13}{4}\right)$
• B. $(-\infty ,-\frac{3}{4})\cup \left(\frac{13}{4},\infty \right)$
• C. $(-\infty ,\frac{1}{4})\cup (-\frac{15}{4},\infty )$
• D. $(-\frac{13}{4},-\frac{3}{4})$

Answer 1

Answer: (B)

$(-\infty ,-\frac{3}{4})\cup \left(\frac{13}{4},\infty \right)$

Given inequality is, $|5-4x|>8$

Case i) Value of x is positive-

$\therefore |5-4x|=5-4x$

$\therefore 5-4x>8$

Adding 4x to both sides, we get,

$5-4x+4x>8+4x$

$\therefore 5>8+4x$

Subtract 8 from both sides,

$\therefore 5-8>8+4x-8$

$\therefore -3>4x$

$\therefore x<\frac{-3}{4}$

Thus, an interval is, $(-\infty ,\frac{-3}{4})$

Case ii) Value of x is negative-

$\therefore |5-4x|=-(5-4x)$

$\therefore |5-4x|=4x-5$

$\therefore 4x-5>8$

Adding 5 to both sides, we get,

$\therefore 4x-5+5>8+5$

$\therefore 4x>13$

$\therefore x>\frac{13}{4}$

Thus, an interval is, $(\frac{13}{4},\infty )$

Thus, combining both the results, we get,

$(-\infty ,\frac{-3}{4})\bigcup (\frac{13}{4},\infty )$

Answer is Option (B)