The following is the equation:
How can I solve this problem?
Do I have to reformat it to $|x+1|=3-|x-2|$?
I would like a simple answer that by no means uses set theory. The answer must include a step by step explanation
You have three cases to look at:
• Case 1: $x<-1$
• Case 2: $-1\le x<2$
• Case 3: $x\ge 2$
The first term $|x+1|$ is the distance from $x$ to $-1$, and the second term $|x-2|$ is the distance from $x$ to $2$. Now, considering that the points $-1$ and $2$ lie precisely $3$ steps apart on the number line, can you deduce where on the the number line $x$ must lie in order for the sum of those two distances to be $3$?
Case 1: $x+1+x+2=3$ >> $x=0$
Case 2: $-(x+1)+x+2=3$ >> No Solutions
Case 3: $x+1-(x+2)=3$ >> No Solutions
Case 4: $-(x+1)-(x+2)=3$ >> $x=-3$
Therefore the two answers are: $x=0$ and $x=-3$
You asked for solutions to $$|x+1|+|x+2|=3.$$
If $x<-2$ then $x+1$ and $x+2 <0$, so this means $-(x+1)-(x+2)=3,$ i.e., $x=-3.$
If $x\ge-1$ then $x+1$ and $x+2 \ge 0$, so this means $(x+1)+(x+2)=3,$ i.e., $x=0$.
If $-2\le x<-1$ then $x+1<0$ but $x+2\ge0$, so $-(x+1)+(x+2)=3$,
which has no solution.
Hint: if $x+1 > 0$ , $x+1 < 0$, $x+1 = 0$ and $x-2 > 0$, $x - 2 < 0$, $x-2 = 0$, etc...There are $9$ possibilities all together.