Solving Modulus Functions

[Srd2000]

Hi Y'all! Is someone able to explain how to solve a modulus function like this example, |x-4|-|x+2|=6 . I understand that you can square both sides, but I'm not that confident with dealing with the modulus functions. Thanks in advance

[keltingmeith]

Hi Y'all! Is someone able to explain how to solve a modulus function like this example, |x-4|-|x+2|=6 . I understand that you can square both sides, but I'm not that confident with dealing with the modulus functions. Thanks in advance

The easiest way (I reckon, people will argue) is to split it up into sections. With two mod functions, you need to go bit by bit. So, we start with one of them:

|x-4|
=x-4 when x-4>=0 ---> x>=4
=-x+4 when x-4<0 ---> x<4

|x+2|
=x+2 when x+2>=0 ---> x>=-2
=-x-2 when x+2<0 ---> x<-2

Okay, so putting those together, we have 4 different scenarios:

1. x>=4 and x>=-2 (x-4 and x+2)
2. x<4 and x>=-2 (-x+4 and x+2)
3. x>=4 and x<-2 (x-4 and -x-2)
4. x<4 and x<-2 (-x+4 and -x-2)

Next, we need to reduce this into different segments where both of those conditions are true.

So, for 1, both situations are true whenever x>=4. For 2, both situations are true for -2<=x<4. For 3, there is no situation where both of those are true. And for 4, both of those are true if x<-2. So, putting that altogether, we have:

1. x-4-(x+2)=6, x>=4
2. -x+4-(x+2)=6, -2<=x<4
4. -x+4-(-x-2)=6, x<-2

Solving these, you'll get:

1. x-4-x-2=-6=6 (Nonsense, ignore this solution)
2. -x+4-x-2=-2x+2=6 ---> -2x=4 ---> x=-2 (this agrees with our bounds, keep it)
4. -x+4+x+2=(x-x)+6=6 (True for any x in this condition - that is, x<-2)

So, the solution is both x=-2 and x<-2 (we can summarise as x<=-2)

If you follow this procedure for every modulus question, you should be fine.

[Srd2000]

Thank you! Are there any other ways of doing it?

[keltingmeith]

Thank you! Are there any other ways of doing it?

As you mentioned, you can square it. On my phone ATM, so I won't go into detail, but:

|x-3|-|x+2|=6
(|x-3|-|x+2|)^2=36
|x-3|^2-2|x-3||x+2|+|x+2|^2=36
-2|(x-3)(x+2)|=36-(x-3)^2-(x+2)^2

Square it again to remove the second modulus. Just beware - you'll always get an extra solution if you square wondering, so you're going to have to check your answers by putting them into the original equation.

There's probably lots of other methods, as well - for example, after squaring the first time, you could solve it as a hybrid function like I did.

[Bri MT]

For this one, you could rearrange to have the mod functions on different sides of the equation & roughly sketch the graphs