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

2 Answers
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 RR => -k < z < k.
By doing this you have that |2x-3| < 5 => - 5 < 2x-3 < 5, which are two inequalities put together. You have to solve them separately:
1st) - 5 < 2x-3 => - 2 < 2x => - 1 < x
2nd) 2x-3 < 5 => 2x < 8 => 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 RR => -k < z < k.
By doing this you have that |2x-3| < 5 => - 5 < 2x-3 < 5, which are two inequalities put together. You have to solve them separately:
1st) - 5 < 2x-3 => - 2 < 2x => - 1 < x
2nd) 2x-3 < 5 => 2x < 8 => x < 4
And, finally, putting both results together (which is always more elegant), you obtain the final result which is - 1 < x < 4.