What is the algebraic proof that $|x|+|x-2|=2$ is an inequality

Question

$|x|+|x-2|=2$ is an inequality because the solution is $x\le2$ (if you plug in $2$ or any value lower than $2$ for $x$, you will get an answer of $2$). What is the algebraic proof (with steps)? How can I write this out using algebra to get a value of $x≤2$?

Answer 1

By the triangle inequality: $$2=|x|+|2-x|\geq|x+2-x|=2.$$ The equality occurs for $x\geq0$ and $2-x\geq0$ or $0\leq x\leq2$, which is the answer.

Answer 2

You have three cases:

1. $x\leq 0$ The equality becomes $-x+2-x=2 \Rightarrow 0=2x \Rightarrow x=0$.
2. $0\leq x\leq 2$ The equality becomes $x+2-x=2 \Rightarrow 2=2$ which is true $\forall x\in[0,2]$.
3. $2\leq x$ The equality becomes $x+x-2=2\Rightarrow 2x=4\Rightarrow x=2$.

In conclusion $x\in[0,2]$ is the set of solutions: $0\leq x\leq 2$.

Answer 3

$y=x-1$;

$|y+1|+|y-1|=2;$

$|2y| =|y+1+y-1| \le $

$|y+1| +|y-1| =2$;

Hence

$|y| \le 1$, i.e. $-1 \le y \le 1$;

With $y=x-1$:

$0 \le x \le 2$.

Answer 4

Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a\;$ and $\;b.$

In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$

The points out the interval $\;[0,2]\;$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $\;|x|+|x-2|=2\;$ can be rewritten $$0\leq x \leq 2.$$

Answer 5

Alternatively, you can square both sides: $$|x|+|x-2|=2 \Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 \Rightarrow \\ |x^2-2x|=-(x^2-2x) \Rightarrow x^2-2x\le 0 \Rightarrow x\in[0,2].$$