Question

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

Answer 1

`(-oo, 3) uu (1, oo)`

Explanation:

Separate the equation into two equations:
`(12-6x)+3>=9`
and
`-(12-6x)+3>=9`
(If you don't understand why, look at the note at the bottom)
Solve both of these equations:
`12-6x+3>=9`
`15-6x>=9`
`-6x>=-6`
`x<=1`

`-12+6x+3>=9`
`-9+6x>=9`
`6x>=18`
`x>=3`
Now, combine the two solutions (OR them):
`(-oo, 3) uu (1, oo)`

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

Answer 2

`x le 1, x ge 3`

Explanation:

We have:

`|12-6x|+3 ge 9 => |12-6x| -6 ge 0`

By definition of the absolute function we have:

`|x| = { (-x, x lt 0), (0, x=0), (x, x gt 0) :}`

And so;

`|12-6x| = { (-(12-6x), 12-6x, lt 0), (0, 12-6x,=0), (12-6x, 12-6x, gt 0) :}`

`" " = { (-12+6x, x gt 2), (0, x=2), (12-x, x lt 2) :}`

Therefore:

`|12-6x| -6 = { (-12+6x-6, x gt 2), (-6, x=2), (12-6x-6, x lt 2) :}`

`" " = { (-18+6x, x gt 2), (-6, x=2), (6-6x, x lt 2) :}`

We require `|12-6x| -6 ge 0`;

Either:

With `x gt 2`
`-18+6x ge 0 => 6x ge 18=> x ge 3`
`:. x in { (x ge 3) uu (x gt 2) } = { x ge 3}`

Or

With `x lt 2`;
`6-6x ge 0 => x le 1`
`:. x in { (x lt 2) uu (x le 1) } = { x le 1}`

So the solution is

`x le 1, x ge 3`