How do you solve #x+ 2\leq 5x - 1\leq x + 3#?

2 Answers
smendyka
Jul 7, 2017

See a solution process below:

Explanation:

First, add #color(red)(1)# to each segment of the system of inequalities to isolate the #x# term in the middle segment while keeping the system balanced:

#x + 2 + color(red)(1) <= 5x - 1 + color(red)(1) <= x + 3 + color(red)(1)#

#x + 3 <= 5x - 0 <= x + 4#

#x + 3 <= 5x <= x + 4#

Next, subtract #color(red)(x)# from each segment to isolate the #x# term while keeping the system balanced:

#-color(red)(x) + x + 3 <= -color(red)(x) + 5x <= -color(red)(x) + x + 4#

#0 + 3 <= -color(red)(1x) + 5x <= 0 + 4#

#3 <= (-color(red)(1) + 5)x <= 4#

#3 <= 4x <= 4#

Now, divide each segment of the system by #color(red)(4)# to solve for #x# while keeping the system balanced:

#3/color(red)(4) <= (4x)/color(red)(4) <= 4/color(red)(4)#

#3/4 <= (color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) <= 1#

#3/4 <= x <= 1#

Or

#x >= 3/4# and #x <= 1#

Or, in interval notation:

#[3/4, 1]#

Binayaka C.
Jul 7, 2017

Solution : # 3/4 <= x <= 1# , in interval notation: # [3/4,1 ]#

Explanation:

# x+2 <= 5x -1 <= x+3 #

1) # x+2 <= 5x -1 or x- 5x <= -1-2 or -4x <= -3 or 4x >= 3 or x >= 3/4 or [3/4,oo)#

2) # 5x -1 <= x+3 or 5x - x <= 3+1 or 4x <= 4 or x <=1 or (-oo , 1]#

The common zone of two solutions is # 3/4 <= x <= 1 or [3/4,1]#

Solution : # 3/4 <= x <= 1# , in interval notation: # [3/4,1 ]# [Ans]

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