Inequalities of the form |A| < k and |A| ≤ k

The right-hand side is the positive constant $2$, so the inequality is equivalent to the double inequality $-2<x-3<2$, written as the system of the two conditions. Adding $3$ to all sides gives the interval $1<x<5$. The geometric reading confirms the result, since $|x-3|<2$ asks for the points at distance less than $2$ from $3$, and these form the open interval with endpoints $1$ and $5$.

The inequality is non-strict, so the two conditions of the system are also non-strict and the endpoints of the interval belong to the solution. In each condition the $5$ is moved to the right and we divide by $2$, which is positive, therefore the direction of the inequalities does not change. Checking at the endpoints confirms the result: for $x=-4$ we have $|-3|=3\le 3$, and for $x=-1$ we have $|3|=3\le 3$; for an exterior point such as $x=0$, instead, we have $|5|=5>3$.

The argument of the absolute value is quadratic, but the rule is the same: the inequality is equivalent to the system of the two conditions $x^2-3\le 1$ and $x^2-3\ge -1$. The condition $x^2\le 4$ is satisfied on the interval $-2\le x\le 2$, while the condition $x^2\ge 2$ is satisfied outside the open interval with endpoints $-\sqrt{2}$ and $\sqrt{2}$. The two conditions must hold together, so we intersect the interval with the union of half-lines, and the two symmetric intervals shown remain. Checking with a test value confirms the result: for $x=\frac{3}{2}$ we have $x^2=\frac{9}{4}$ and $\left|\frac{9}{4}-3\right|=\frac{3}{4}\le 1$, while for $x=0$ we have $|-3|=3>1$.