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Question

Find the values of $x$ such that $|x-2|=2-x$. How do I solve this?

Answer 1

Let $t=x-2$. Then $|t|=-t$ is true for non-positive $t$. Hence $x-2\le0$.

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More generally, and equation like

$$f(x)=\sum_k|a_kx+b_k|+ax+b=0$$

is piecewise linear, and it suffices to observe changes of sign between two "corner points", which occur where $x=-\dfrac{b_k}{a_k}$. You sort these points increasingly, including $-\infty$ and $\infty$ to cover the whole real axis, and evaluate the $f(x_k)$.

Here, with $f(x):=|x-2|+x-2$, the only corner point occurs at $x=2$, with $f(2)=0$. We have

$$f(-\infty)=0,f(2)=0,f(\infty)=\infty$$ so that $f(x)=0$ for $x\in (-\infty,2]$.

Answer 2

For $$x\geq 2$$ we get $$x-2=2-x$$ so $x=2$. For $x<2$ we get $$-x+2=2-x$$ this is true for all $x$ with $x<2$.

Answer 3

As $|a|=\max(a,-a)\ge 0$, this means that $2-x\ge 0$, i.e. $x\le 2$.

Answer 4

As $|x-2|\ge 0$, we must have $2-x\ge0$. And for all $x$ that satisfy this condition, we do have $|2-x|=2-x=|x-2|$.