How do you solve this simple equality containing a modulo operator

[itsrayex]

Hi, I have the following equation to solve:

A very reliable website that I am learning from suggests ways to solve these kind of equalities:


But it seems to me that this equality is of the kind "p - |A(x)| = k". Sure, I could move the p to the right so it would be an equality of the first type, but my issue is that I don't know how to deal with that minus at the beginning, and the website does not provide any info about such cases. Can you please tell me how to solve this kind of equality?
Also, the website suggests ways to solve equalities of the types I mentioned above without having to put the equalities into systems with inequalities (something I have vague memories of from high school), which I appreciate very much. I'm afraid it will not be the same for this case, right? There's no workaround and I must use inequalities somehow right, because of the minus at the beginning?
These are two attempts of me trying to solve the equality with the knowledge that I have right now. I had made a third attempt but deleted it:

Thank you to anyone who will help!

 

[Zermelo]

So by definition, |x| = x, when x is non-negative (x>=0), and |x| = - x, when x is negative (x<0)
In the same way |2-x| = 2-x, when 2-x >= 0, and |2-x| = x - 2 when 2 - x < 0.
You can consider each of these cases 2-x >= 0 and 2 - x < 0 seperately, and for each you need to solve a simple linear equation without | |. I think this is the most intuitive way to do this, don’t go memorizing formulas if you don’t know what they mean or why are they true, you’re just going to forget them in 2 days ?
But if you really want to do it that way, move the |2-x| term to the right hand side, and 3 to the left, and then you will have the shape you are looking for! 7-3 = |2-x|

 

[BigBeachBanana]

By modulo, I think you meant absolute values. In general, [math]|x| = \begin{cases} -x \\\, x\\ \end{cases}[/math].
So we have
[math]7-|2-x| = \begin{cases} 7--(2-x)\\ 7- (2-x) \end{cases}= \begin{cases} 7+(2-x)\\ 7- (2-x) \end{cases}[/math]Lastly, set them equal to 3 and solve.
[math]\begin{cases} 7+(2-x)=3\\ 7- (2-x)=3 \end{cases}[/math]

 

[Dr.Peterson]

Hi, I have the following equation to solve:
View attachment 30852
A very reliable website that I am learning from suggests ways to solve these kind of equalities:

View attachment 30850
But it seems to me that this equality is of the kind "p - |A(x)| = k". Sure, I could move the p to the right so it would be an equality of the first type, but my issue is that I don't know how to deal with that minus at the beginning, and the website does not provide any info about such cases. Can you please tell me how to solve this kind of equality?
Also, the website suggests ways to solve equalities of the types I mentioned above without having to put the equalities into systems with inequalities (something I have vague memories of from high school), which I appreciate very much. I'm afraid it will not be the same for this case, right? There's no workaround and I must use inequalities somehow right, because of the minus at the beginning?
These are two attempts of me trying to solve the equality with the knowledge that I have right now. I had made a third attempt but deleted it:
View attachment 30851
Thank you to anyone who will help!

You need to know that this does not follow:

If you think you are following a formula there, you are not. It is only when one side is only an absolute value that you can drop the absolute values and insert signs on the other side.

To solve this by that method, you need to first isolate the absolute value, so that it takes one of the forms you listed. The formulaic methods apply only to those specific forms, not to anything that has an absolute value somewhere. (And I'm sure that your source says this, and perhaps includes examples where they put an equation into the right form.)

In particular, [imath]|A|=k[/imath] can be rewritten as [imath]A=\pm k[/imath] for a reason: the two numbers whose absolute values are k are +k and -k (assuming k is positive). That reason does not apply otherwise.

 

[pka]

Hi, I have the following equation to solve:
View attachment 30852

Do you understand that [imath]|2-x|=|x-2|~?[/imath] If so
then we can rewrite the question [imath](7-3)-|x-2|=0[/imath].
[imath]\text{Or }4-|x-2|=0[/imath] From which we can easily see that [imath]x=-2\text{ or }x=6.[/imath]
[imath][/imath][imath][/imath]

 

[Steven G]

7 - |2-x| = 3
It should be obvious that |2-x|= 4. After all 7 - 4 = 3
|2 - x| = 4 means the distance between x and 2 is 4. So x can be 2-4 or x can be 2+4. So x= -2 or x = 6