You are Here: Home >< Maths

# Maths Modulus Question: |x-a| + |x+a| = 2a, where a > 0. watch

1. Hi, I have tried to solve this question for a very long time now, but still cannot get to the solution. I have divided the problem into 3 separate cases of:
1) x is greater than a.
2) x is smaller than -a.
3) x is greater than or equal to -a and smaller than or equal to a.

But I still can't get anywhere.

Please could you provide a solution to this question.
Hi, I have tried to solve this question for a very long time now, but still cannot get to the solution. I have divided the problem into 3 separate cases of:
1) x is greater than a.
2) x is smaller than -a.
3) x is greater than or equal to -a and smaller than or equal to a.

But I still can't get anywhere.

Please could you provide a solution to this question.
Can't you remove the modulus-es differently for each according to these separate three cases, and then find solutions for each?
3. (Original post by AHappyStudent)
Can't you remove the modulus-es differently for each according to these separate three cases, and then find solutions for each?
I have done that and I got x = a, x = -a and a = a
4. Then for each of these solutions shouldn't you test them against their original bounds to see if each one is valid?

For x=a, is this valid as the condition on which it is based was x > a ? (I don't think so)

For x = -a, is this valid as the condition on which it was based was x < -a? (I don't think so)

For a = a, x can be any value as long as it lies in the range -a <= x <= a ?

I'm not sure if this solution is correct
5. (Original post by AHappyStudent)
Then for each of these solutions shouldn't you test them against their original bounds to see if each one is valid?

For x=a, is this valid as the condition on which it is based was x > a ? (I don't think so)

For x = -a, is this valid as the condition on which it was based was x < -a? (I don't think so)

For a = a, x can be any value as long as it lies in the range -a <= x <= a ?

I'm not sure if this solution is correct
THANKS! That is the solution in the textbook! This explanation is really clear. Thanks a lot for your help.
THANKS! That is the solution in the textbook! This explanation is really clear. Thanks a lot for your help.
Ooh, that's quite a nice little solution then. np, happy to help

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 7, 2017
Today on TSR

### Uni league tables

Do they actually matter?

### University open days

• University of Exeter
Wed, 24 Oct '18
Wed, 24 Oct '18
• Northumbria University
Wed, 24 Oct '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams