How do you solve  5 |x - 3| >=20?

$x \ge 7 \mathmr{and} x \le - 1$. In interval notation, solution is $\left(- \infty , - 1\right] \cup \left[7 , + \infty\right)$
$5 | x - 3 | \ge 20 \mathmr{and} | x - 3 | \ge 4 \therefore x - 3 \ge 4 \mathmr{and} x \ge 7$ OR
$5 | x - 3 | \ge 20 \mathmr{and} | x - 3 | \ge 4 \therefore x - 3 \le - 4 \mathmr{and} x \le - 1$
In interval notation solution is $\left(- \infty , - 1\right] \cup \left[7 , + \infty\right)$[Ans]