
Module Equations.
Greetings,
I'm new to this forum and I require urgent help. I was gone from my country for 2 weeks and meanwhile my class revised the math stuff from previous year. Can anyone help?
x3=1
x+4=2
x=2+x
x2=2
Sorry if my english isn't pleasing, it's not my first language. Thanks indeed.

Re: Module Equations.
Sure. I'll refer to these "modules" as "absolute values".
The key thing to remember is that, for instance,
$\displaystyle 3 = 3$ and
$\displaystyle 3 = 3$
In other words, if the number inside the lines is negative, it becomes positive, but if it's positive, it stays positive. So each of these is two problems in one.
Your first problem is a trick question; an absolute value cannot be negative.
Your second problem, $\displaystyle x+4=2$, can be solved. If $\displaystyle x+4$ is positive, then its absolute value remains x+4, so we have:
$\displaystyle x+4=2$
$\displaystyle x=2$
However, if x+4 is negative, then its absolute value will become positive. This means that it is multiplied by 1. So we have
$\displaystyle (x+4)=$
$\displaystyle x4=2$
$\displaystyle x=6$
$\displaystyle x=6$
Thus the two solutions for the second problem are $\displaystyle 2$ and $\displaystyle 6$.

Re: Module Equations.
The third problem works in the same way, but with a twist.
$\displaystyle x=2+x$
If x is positive, then its absolute value is also positive. So we have:
$\displaystyle x=2+x$, and by subtracting $\displaystyle x$ from both sides we get
$\displaystyle 0=2$, which is absurd. Thus there is no answer that makes the equation true if $\displaystyle x$ is positive.
Suppose x was negative. Then its absolute value would be positive, which means it would be x. In that case,
$\displaystyle x=2+x$
$\displaystyle 2x=2$
$\displaystyle x=1$
So there is only one solution in this case: $\displaystyle 1$.

Re: Module Equations.
The fourth problem involves this process used twice (an absolute value within an absolute value)

Firstly, we take the innermost absolute value. If $\displaystyle x$ is positive, then we have:
$\displaystyle x2=2$
From here, if $\displaystyle x2$ is positive, we have:
$\displaystyle x2=2$
$\displaystyle x=4$
And if $\displaystyle x2$ is negative, then $\displaystyle (x2)=x+2$ is positive, so
$\displaystyle x+2=2$
$\displaystyle x=0$.

Now suppose $\displaystyle x$ is negative. Then $\displaystyle x$ is positive. In that case,
$\displaystyle x2=2$
If $\displaystyle x2$ is positive, we have:
$\displaystyle x2=2$
$\displaystyle x=4 $
$\displaystyle x=4$
If $\displaystyle x2$ is negative, then $\displaystyle (x2)=x+2$ is positive, and we have:
$\displaystyle x+2=2$
$\displaystyle x=0$
This gets us through all the possibilities. Thus the possible answers are 4, 0, and 4.