Question

Solve the following linear in-equation and graph the solution set on a real number line:

${\displaystyle \frac{5}{4}\text{x}>1+\frac{1}{3}(4\text{x}-1)}$ , x ∈ R

Answer 1

${\displaystyle \frac{5}{4}\text{x}>1+\frac{1}{3}(4\text{x}-1)}$

${\displaystyle \frac{5}{4}\text{x}>\frac{3+(4\text{x}-1)}{3}}$

15x > 12 + 16x - 4

15x - 16x > 8

- x > 8

x < -8

Solution set = [x < -8]