How do you solve #-8\leq 8x - 3< 2#?

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Answer 1 · 2017-02-20T23:29:01

See the entire solution process below:

Step 1) Add #color(red)(3)# to each segment of the system of inequalities to isolate the #x# term while keeping the system in balance:

#-8 + color(red)(3) <= 8x - 3 + color(red)(3) < 2 + color(red)(3)#

#-5 <= 8x - 0 < 5#

#-5 <= 8x < 5#

Step 2) Divide each segment of the system of inequalities by #color(red)(8)# to solve for #x# while keeping the system in balance:

#-5/color(red)(8) <= (8x)/color(red)(8) < 5/color(red)(8)#

#-5/8 <= (color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) < 5/8#

#-5/8 <= x < 5/8#