How do you solve #-4( x - 1) \leq - 3( x - 5)#?

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Answer 1 · 2016-09-24T13:37:28

#-11 <= x#

First we expand the brackets

#-4(x-1) <= -3(x-5) to #

#-4x + 4 <= -3x + 15#

Then we add #4x# to both sides

#4 <= x + 15 #

Finally we take #15# from both sides

#-11 <= x#