How do you solve and write the following in interval notation: #abs(3(x+2)-7x)<=6#?

1 Answer
Oct 16, 2016

Answer:

#0<=x<=3#

Explanation:

#∣3x+6-7x∣<=6#
#∣6-4x∣<=6#
#6-4x<=6# and #-(6-4x)<=6#
#-4x<=0# and #-6+4x<=6#
#4x>=0# and #4x<=12#
#x>=0# and #x<=3#
the answer is #0<=x<=3#
Checking, let #x=2#
LHS #6-4*2=-2# and #∣-2∣=2#
RHS #=6#
so #2# is #<=6#