Some confusion with absolute value

Question

Today at a math lecture, I solved the equation $|x+1|+|x-1|+|x|=4$ by using elementary arithmetic.

But my professor did it a little bit differently:
I didn`t pay attention to the teacher's method,because I thought that I could understand it, but no, I couldn't.

I do not understand what does pluses and minuses mean.
Might someone enlighten me?

Answer 1

$$|x+1|+|x-1|+|x|=4$$ then $x+1=0\Rightarrow x=-11$ for $x>-1$ you have $x-1>0$ so $|x+1|=x+1$ and for $x\leq -1$ $|x+1|=-(x+1)$

Using $$|x+1|+|x-1|+|x| = \left\{\begin{matrix} - (x+1)-(x-1)-x&\;, x\leq -1 \\\\ x+1-(x-1)-x & \;, -1<x \leq 0\\\\ x+1-(x-1)+x& , 0<x \leq 1\\\\ x+1+(x-1)+x& , x>1 \end{matrix}\right.$$

$\bullet\; $ If $x\leq -1\;,$ Then equation convert into $- (x+1)-(x-1)-x=-3x=4\Rightarrow x=\frac{-4}{3}\;\; \text{(true)}$

$\bullet\; $ If $-1<x \leq 0\;,$ Then equation convert into $x+1-(x-1)-x=4\Rightarrow \displaystyle x={2}\;\; \text{(False)}$

$\bullet\; $ If $x\geq 1\;,$ Then equation convert into $x+1+(x-1)+x=4\Rightarrow x=\frac{4}{3}\;\; \text{(True)}$

Answer 2

Your professor was giving the 4 cases of what signs the 3 absolute value terms would take depending on where $x$ is, either $x \geq 1$ or $0 \leq x \leq 1$ or $-1 \leq x \leq 0$ or $x \leq -1$. You can solve the equation for $x$ in each of these 4 cases because you can drop the absolute values and multiply each term by either $+1$ or $-1$, and then make sure the value of $x$ you get when you solve is within the range it is supposed to be. Whichever cases make this happen, this gives you your valid solutions for $x$.

Answer 3

Your diagram shows the signs of the three arguments of the absolute value expressions in the intervals of the real line delimited by the critical points $x=-1$, $x=0$ and $x=1$ ( the points where the arguments are zero).

By definition, where an argument is negative the absolute value is the opposite of the argument and where the argument is positive, the absolute value is the same as the argument. So, es an example, for $x<-1$ you have to change all the signs of the three arguments and you find: $$ x<-1 \rightarrow -(x+1)-(x-1)-x=4 $$

Doing the same for the four intervals you find four equations that have to be verified for $x$ in the given interval, and the union of all these solutions is the solution of the starting equation.