How do you solve #-8<= (- 3x + 1)/4 <= 15#?

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Answer 1 · 2017-07-13T02:05:22

Given: #-8<= (- 3x + 1)/4 <= 15" [1]"#

Multiply everything by 4:

#-32 <= -3x+1 <= 60#

Subtract 1 from everything:

#-33 <= -3x <= 59#

Divide everything by -3:

#11 >= x >= -59/3#

Flip it so that the negative number is no the left:

#-59/3 <= x <=11#

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