How do you solve #abs(2x-3)>1#?

1 Answer
Oct 7, 2015

Answer:

#x < 1 " or " x > 2#

Explanation:

If #x >= 3/2#
then #abs(2x-3) = 2x-3#
#color(white)("XX")abs(2x-3) > 1#
#color(white)("XX")rarr 2x-3 >1#
#color(white)("XX")rarr 2x > 4#
#color(white)("XX")rarr x > 2#

If #x < 3/2#
then #abs(2x-3) = 3-2x#
#color(white)("XX")abs(2x-3) > 1#
#color(white)("XX")rarr 3-2x > 1#
#color(white)("XX")rarr 3 > 2x+1#
#color(white)("XX")rarr 1 > 2x#
#color(white)("XX")rarr 1 > x#

Combining the two case we get
#color(white)("XX")x < 1 " or " x > 2#