How do you solve #(1/4)(3x + 2) ≤ (2/3)(-3 + 3x)#?

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Answer 1 · 2015-05-12T20:21:14

First notice that the right hand side can be rearranged as follows:

#(2/3)(-3+3x)#

#=(2/3)(3*(-1)+3*x)#

#=(2/3)*3(-1+x)#

#=2(x - 1)#

#=2x-2#

So we have:

#(1/4)(3x + 2) <= 2x - 2#

Multiply both sides by 4 to get:

#3x + 2 <= 8x - 8#

Add 8 to both sides to get:

#3x + 10 <= 8x#

Subtract #3x# from both sides to get:

#10 <= 5x#

Divide both sides by 5 to get:

#2 <= x#

Or to put it another way:

#x >= 2#