# How do you solve -2abs(x-3)+6 < -4?

May 16, 2015

There are two possibilities: $x - 3$ (without the absolutes) is positive or negative (the 'switch point is $x = 3$)

If it is positive, we can leave out the absolute brackets, is it is negative, we 'turn around' to $- \left(x - 3\right) = 3 - x$

(1) $x - 3 \ge 0 \to x \ge 3$
$- 2 \left(x - 3\right) + 6 < - 4 \to - 2 x + 6 + 6 < - 4 \to$
$- 2 x < - 16 \to x > 8$

(2)$x - 3 < 0 \to x < 3$
$- 2 \left(3 - x\right) + 6 < - 4 \to - 6 + 2 x + 6 < - 4 \to$
$2 x < - 4 \to x < - 2$

Conclusion: $x < - 2 \mathmr{and} x > 8$