# How do you solve the absolute value inequality abs(2x - 3)< 5?

Apr 9, 2015

The result is $- 1 < x < 4$.

The explanation is the following:

In order to be able to supress the absolute value (which is always disturbing), you can apply the rule: $| z | < k , k \in \mathbb{R} \implies - k < z < k$.
By doing this you have that $| 2 x - 3 | < 5 \implies - 5 < 2 x - 3 < 5$, which are two inequalities put together. You have to solve them separately:
1st) $- 5 < 2 x - 3 \implies - 2 < 2 x \implies - 1 < x$
2nd) $2 x - 3 < 5 \implies 2 x < 8 \implies x < 4$
And, finally, putting both results together (which is always more elegant), you obtain the final result which is $- 1 < x < 4$.

Apr 9, 2015

The result is $- 1 < x < 4$.

The explanation is the following:

In order to be able to supress the absolute value (which is always disturbing), you can apply the rule: $| z | < k , k \in \mathbb{R} \implies - k < z < k$.
By doing this you have that $| 2 x - 3 | < 5 \implies - 5 < 2 x - 3 < 5$, which are two inequalities put together. You have to solve them separately:
1st) $- 5 < 2 x - 3 \implies - 2 < 2 x \implies - 1 < x$
2nd) $2 x - 3 < 5 \implies 2 x < 8 \implies x < 4$
And, finally, putting both results together (which is always more elegant), you obtain the final result which is $- 1 < x < 4$.