How do you solve # 12x < 22+ x#?

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Answer 1 · 2017-07-21T02:51:51

See a solution process below:

First, subtract #color(red)(x)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#12x - color(red)(x) < 22 + x - color(red)(x)#

#12x - color(red)(1x) < 22 + 0#

#(12 - color(red)(1))x < 22#

#11x < 22#

Now, divide each side of the inequality by #color(red)(11)# to solve for #x# while keeping the inequality balanced:

#(11x)/color(red)(11) < 22/color(red)(11)#

#(color(red)(cancel(color(black)(11)))x)/cancel(color(red)(11)) < 2#

#x < 2#