Find the values of $x$ such that $|x-2|=2-x$. How do I solve this?
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3I would suggest you start by searching for the definition of absolute value. – Arnaud D. Sep 22 '19 at 12:56
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Note that $2-x=-(x-2)$. Hence, this is equivalent to $x-2\leq0$. – Rushabh Mehta Sep 22 '19 at 12:58
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How is x≤2 is true for every value of x? What principle is this? – Pisting Inatay Sep 22 '19 at 13:06
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By the definition of absolute value $|x-2| = -(x-2)$ when $x < 2$, and $|x-2| = x-2$ when $x > 2$. This is because only when $x-2<0$ does the absolute sign take effect and make the LHS positive. – Toby Mak Sep 22 '19 at 13:08
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Okay, took myself some time. But this really helped. Thank you! – Pisting Inatay Sep 22 '19 at 13:21
4 Answers
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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 $x<2$.
Dr. Sonnhard Graubner
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Let $t=x-2$. Then $|t|=-t$ is true for non-positive $t$. Hence $x-2\le0$.
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 $x\in (-\infty,2]$.
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As $|x-2|\ge 0$, we must have $2-x\ge0$. And for all $x$ that satisfy this condition, we do have $|2-x|=2-x=|x-2|$.