# How do you solve |12- 6x | + 3\geq 9?

Mar 20, 2017

$\left(- \infty , 3\right) \cup \left(1 , \infty\right)$

#### Explanation:

Separate the equation into two equations:
$\left(12 - 6 x\right) + 3 \ge 9$
and
$- \left(12 - 6 x\right) + 3 \ge 9$
(If you don't understand why, look at the note at the bottom)
Solve both of these equations:
$12 - 6 x + 3 \ge 9$
$15 - 6 x \ge 9$
$- 6 x \ge - 6$
$x \le 1$

$- 12 + 6 x + 3 \ge 9$
$- 9 + 6 x \ge 9$
$6 x \ge 18$
$x \ge 3$
Now, combine the two solutions (OR them):
$\left(- \infty , 3\right) \cup \left(1 , \infty\right)$

NOTES:
So, this is a standard equation $y \le \left\mid x \right\mid$:
graph{y<=abs(x) [-5, 5, -5, 5]}
This graph is sort of a piecewise function of the two linear graphs of $\le x$:
graph{y<=x [-5, 5, -5, 5]}
and $y \le - x$:
graph{y<=-x [-5, 5, -5, 5]}
And the solution to the first equation is actually ta combination of the two inequalities.

Mar 20, 2017

$x \le 1 , x \ge 3$

#### Explanation:

We have:

$| 12 - 6 x | + 3 \ge 9 \implies | 12 - 6 x | - 6 \ge 0$

By definition of the absolute function we have:

$| x | = \left\{\begin{matrix}- x & x < 0 \\ 0 & x = 0 \\ x & x > 0\end{matrix}\right.$

And so;

$| 12 - 6 x | = \left\{\begin{matrix}- \left(12 - 6 x\right) & 12 - 6 x & < 0 \\ 0 & 12 - 6 x & = 0 \\ 12 - 6 x & 12 - 6 x & > 0\end{matrix}\right.$

$\text{ } = \left\{\begin{matrix}- 12 + 6 x & x > 2 \\ 0 & x = 2 \\ 12 - x & x < 2\end{matrix}\right.$

Therefore:

$| 12 - 6 x | - 6 = \left\{\begin{matrix}- 12 + 6 x - 6 & x > 2 \\ - 6 & x = 2 \\ 12 - 6 x - 6 & x < 2\end{matrix}\right.$

$\text{ } = \left\{\begin{matrix}- 18 + 6 x & x > 2 \\ - 6 & x = 2 \\ 6 - 6 x & x < 2\end{matrix}\right.$

We require $| 12 - 6 x | - 6 \ge 0$;

Either:

With $x > 2$
$- 18 + 6 x \ge 0 \implies 6 x \ge 18 \implies x \ge 3$
$\therefore x \in \left\{\left(x \ge 3\right) \cup \left(x > 2\right)\right\} = \left\{x \ge 3\right\}$

Or

With $x < 2$;
$6 - 6 x \ge 0 \implies x \le 1$
$\therefore x \in \left\{\left(x < 2\right) \cup \left(x \le 1\right)\right\} = \left\{x \le 1\right\}$

So the solution is

$x \le 1 , x \ge 3$