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Answer 1 · 2016-09-24T00:06:02

$x in [2, oo)$

The absolute value function is defined as

$|a| = {(a" if "a>=0), (-a" if "a<0):}$

Thus, we consider two cases:

Case 1: $x-2 >= 0$

Then $|x-2| = x-2$ , so

$3(x-2) = 3x-6$

$=> 3x-6 = 3x-6$

As this is a tautology, the equation is true for any $x$ satisfying $x-2 >= 0$ ... that is, for any $x >= 2$

Case 2: $x-2 < 0$

Then $|x-2| = -(x-2) = 2-x$ , so

$3(2-x) = 3x-6$

$=> 6-3x = 3x-6$

$=> 6x = 12$

$=> x = 2$

As we began this case with the restriction $x-2 < 0$ , that is, $x < 2$ , there are no solutions within the interval $(-oo, 2)$ .

Taken together, we can see that the equation holds true if and only if $x >= 2$ , so we have the solution set $x in [2, oo)$ .