How do you solve and write the following in interval notation: #|2x + 3| < 11#?

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Answer 1 · 2018-04-02T06:38:18

The solution is #x in(-7,4)#

There are #2# solutions for equation with absolute values.

#|2x+3|<11#

#2x+3<11# and #-2x-3<11#

#2x<8# and #2x > -14#

#x<4# and #x > -7#

The solutions are

#S_1=x in( -oo, 4)#

#S_2=x in (-7,+oo)#

Therefore,

#S=S_1nnS_2#

#=x in ( -7, 4)#

graph{(y-|2x+3|)(y-11)=0 [-5.55, 6.934, -0.245, 5.995]}