Solve$\left|x-4\right|<2$

$\left|x-4\right|<2$

The solution to the inequality $\left|x\right|<a$is $x<a$$x>-a$and $x<a$

That implies the solution of the inequality $\left|x\right|<a$is the intersection of the solutions of the inequalities $x>-a$ and $x<a$.

The solution to the given inequality $\left|x-4\right|<2$is

Case I: $x-4$is non-negative.

$\begin{array}{c}x-4<2\\ x-4+4<2+4\\ x<6\\ x\in \left(-\infty ,6\right)\end{array}$

Case 2: $x-4$is negative.

$\begin{array}{c}-\left(x-4\right)<2\\ \left(-1\right)\left(-\left(x-4\right)\right)>\left(-1\right)\left(2\right)\\ x-4>-2\\ x-4+4>-2+4\\ x>2\\ x\in \left(2,\infty \right)\end{array}$

The solution to the inequality$\left|x\right|<a$is $x>-a$and $x<a$

That implies the solution of the inequality $\left|x\right|<a$is the intersection of the solutions of the inequalities $x>-a$and $x<a$.

Find the intersection of the solutions of the inequalities $x-4<2$and $x-4>-2$find the solution of the inequality $\left|x-4\right|<2$.

The intersection of the solutions of the inequalities $x-4<2$and $x-4>-2$is : $x\in \left(2,6\right)$

Therefore, the solution to the inequality $\left|x-4\right|<2$is $x\in \left(2,6\right)$.