How do you solve #|12- 6x | + 3\geq 9#?
2 Answers
Explanation:
Separate the equation into two equations:
and
(If you don't understand why, look at the note at the bottom)
Solve both of these equations:
Now, combine the two solutions (OR them):
NOTES:
So, this is a standard equation
graph{y<=abs(x) [-5, 5, -5, 5]}
This graph is sort of a piecewise function of the two linear graphs of
graph{y<=x [-5, 5, -5, 5]}
and
graph{y<=-x [-5, 5, -5, 5]}
And the solution to the first equation is actually ta combination of the two inequalities.
# 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
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 #