How do you solve the compound inequality #3<2x-3<15#?

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Answer 1 · 2017-05-21T11:33:12

See a solution process below:

First, add #color(red)(3)# to each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#3 + color(red)(3) < 2x - 3 + color(red)(3) < 15 + color(red)(3)#

#6 < 2x - 0 < 18#

#6 < 2x < 18#

Now, divide each segment by #color(red)(2)# to solve for #x# while keeping the system balanced:

#6/color(red)(2) < (2x)/color(red)(2) < 18/color(red)(2)#

#3 < (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) < 9#

#3 < x < 9#

Or

#x > 3#; #x < 9#

Or, in interval notation:

#(3, 9)#

Answer 2 · 2017-05-21T11:34:18

Solution: # 3 < x < 9 # or in interval notation: #(3,9) #

# 3 < 2x-3 <15 # Adding #3# in all sides we get

# 6 < 2x <18 # Multiplying by #1/2# in all sides we get

# 3 < x < 9 #.

Solution: # 3 < x < 9 # or in interval notation: #(3,9) # [Ans]