Intermediate Algebra (6th Edition)

The solution is $\Big(-\infty,-\dfrac{8}{3}\Big]\cup(2,\infty)$
$\dfrac{3}{8}x+1\le0$ or $-2x\lt-4$ Solve the first inequality: $\dfrac{3}{8}x+1\le0$ Take $1$ to the right side: $\dfrac{3}{8}x\le-1$ Take $8$ to multiply the right side: $3x\le-8$ Take $3$ to divide the right side: $x\le-\dfrac{8}{3}$ Expressing the solution in interval notation: $\Big(-\infty,-\dfrac{8}{3}\Big]$ Solve the second inequality: $-2x\lt-4$ Take $2x$ to the rigth side and $4$ to the left side: $4\lt2x$ Rearrange: $2x\gt4$ Take $2$ to divide the right side: $x\gt\dfrac{4}{2}$ $x\gt2$ Expressing the solution in interval notation: $(2,\infty)$ Since the compound inequality is formed by the word "or", the solution is composed by the numbers that satisfy each inequality: The solution is $\Big(-\infty,-\dfrac{8}{3}\Big]\cup(2,\infty)$