How do you solve #16\leq 2y + 20\leq 20#?

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Answer 1 · 2017-11-19T13:04:34

See a solution process below:

First, subtract #color(red)(20)# from each segment of the system of inequalities to isolate the #y# term while keeping the system balanced:

#16 - color(red)(20) <= 2y + 20 - color(red)(20) <= 20 - color(red)(20)#

#-4 <= 2y + 0 <= 0#

#-4 <= 2y <= 0#

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

#-4/color(red)(2) <= (2y)/color(red)(2) <= 0/color(red)(2)#

#-2 <= (color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) <= 0#

#-2 <= y <= 0#

Or

#y >= -2# and #y <= 0#

Or, in interval notation

#[-2, 0]#