Answer
$$x\geq \frac{4a+ac-d}{ab}$$
OR
$$x\leq \frac{-4a+ac+d}{ab}$$
Work Step by Step
$$a|bx-c|+d\geq 4a$$
To solve this inequality, we have to consider two possible way. That is when value of $|bx-c|$ is positive and negative.
$$a\times ±(bx-c)+d\geq 4a$$
So we have:
$$a(bx-c)+d\geq 4a$$ $$abx-ac+d\geq 4a$$ $$abx\geq 4a+ac-d$$ $$x\geq \frac{4a+ac-d}{ab}$$
OR
$$-a(bx-c)+d\geq 4a$$ $$-abx+ac+d\geq 4a$$ $$-abx\geq 4a-ac-d$$ $$x\leq \frac{-4a+ac+d}{ab}$$
