How do you solve and graph abs(k-4)>3?

Dec 26, 2017

$k < 1$ or $k > 7$

Explanation:

$\left\mid k - 4 \right\mid > 3$
when does $\left\mid k - 4 \right\mid = 3$?
$\left\mid k - 4 \right\mid = 3 \iff k - 4 = 3 \mathmr{and} k - 4 = - 3 \iff k = 7 \mathmr{and} k = 1$

Now let's check what happens:
when $k < 1$
$\left\mid 0 - 4 \right\mid = 4 > 3$
when $k > 7$
$\left\mid 10 - 4 \right\mid = 6 > 3$
when $1 > k > 7$
$\left\mid 5 - 4 \right\mid = 1 < 3$

so we need $k < 1$ or $k > 7$