How do you solve #2x-3<x+4<3x-2#?

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Answer 1 · 2015-07-27T21:20:10

#3 < x < 7# (In interval notation: #(3, 7)#)

In order for #2x-3 < x+4 < 3x-2# to be true, the compound inequality:

#2x-3 < x+4# #" "# AND #" "# #x+4 < 3x-2# must be true.

Solving each individually, we get:

#2x-3 < x+4# Subtract #x# and add #3# on both sides to get:

#x < 7#

We also need:

#x+4 < 3x-2# Subtract #x# and add #2# on both sides to get:

#6 < 2x#, so #3 < x#

The solution will require both #3 < x# and #x < 7#.

The numbers that satisfy both inequalities simultaneously are the numbers between #3# and #7# (exclusive).