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

2 Answers
Mar 20, 2017

Answer:

#(-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.

Mar 20, 2017

Answer:

# 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 #