Answer
The solution is $\Big(-\infty,-\dfrac{8}{3}\Big]\cup(2,\infty)$
Work Step by Step
$\dfrac{3}{8}x+1\le0$ 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 $\Big(-\infty,-\dfrac{8}{3}\Big]\cup(2,\infty)$
