Hello, struck!
Solve: .$\displaystyle x1 \:=\:x + 1$ I'll do it the Long Way . . .
Since both $\displaystyle x1$ and $\displaystyle x$ can be positive or negative,
. . there are four cases to consider:
[1] Both positive: .$\displaystyle x1 > 0,\;\;x> 0 \quad\Rightarrow\quad x > 1$
. . .Then we have: .$\displaystyle x  1 \:=\:x + 1 \quad\Rightarrow\quad 0 \:=\:2$ . . . impossible
[2] Positivenegative: .$\displaystyle x  1 > 0,\;\;x < 0$
. . .But this means: .$\displaystyle (x > 1) \:\wedge\:(x < 0)$ . . . impossible
[3] Negativepositive: .$\displaystyle x1 < 0,\;\;x > 0$
. . .Then we have: .$\displaystyle 1  x \:=\:x + 1 \quad\Rightarrow\quad x = 0 $
[4] Negativenegative: .$\displaystyle x1 < 0,\;\;x < 0 \quad\Rightarrow\quad x < 0$
. . .Then we have: .$\displaystyle 1  x \:=\:x + 1 \quad\Rightarrow\quad 0 \:=\:0$ . . . always true.
The inequality is satisfied in cases [3] and [4]
The solution is: .$\displaystyle x \:\leq \:0$
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If we graph the two functions, we can "eyeball" the solution.
The graph of $\displaystyle y = x$ is a $\displaystyle \vee$, vertex at (0,0). .[1]
The graph of .$\displaystyle y \:=\:x1$ . is [1], moved one unit to the right. Code:
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1* /
\ /
 \ /
 \ /
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    +   *    
 1
The graph of .$\displaystyle y \:=\:x+1$ .is [1], moved one unit upward. Code:

\  /
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\  /
\  /
\/
1*





    +       

Sketch them on the same coordinate system
. . and we can see the solotion: .$\displaystyle x \:\leq\:0$