How do you solve #2(x-3)<3(2x+2)#?

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Answer 1 · 2017-02-18T22:34:00

First, expand the terms within parenthesis by multiplying each term with the parenthesis by the term outside the parenthesis:

#(color(red)(2) xx x) - (color(red)(2) xx 3) < (color(red)(3) xx 2x) + (color(red)(3) xx 2)#

#2x - 6 < 6x + 6#

Next, subtract #color(red)(2x)# and #color(blue)(6)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#2x - 6 - color(red)(2x) - color(blue)(6) < 6x + 6 - color(red)(2x) - color(blue)(6)#

#2x - color(red)(2x) - 6 - color(blue)(6) < 6x - color(red)(2x) + 6 - color(blue)(6)#

#0 - 12 < 4x + 0#

#-12 < 4x#

Now, divide each side of the inequality by #color(red)(4)# to solve for #x# while keeping the inequality balanced:

#(-12)/color(red)(4) < (4x)/color(red)(4)#

#-3 < (color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4))#

#-3 < x#

To put the solution in terms of #x# we can reverse or "flip" the inequality:

#x > -3#