How do you solve #x+ 8< 3x + 4#?

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Answer 1 · 2017-06-13T21:20:41

See a solution process below:

First, subtract #color(red)(x)# and #color(blue)(4)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#-color(red)(x) + x + 8 - color(blue)(4) < -color(red)(x) + 3x + 4 - color(blue)(4)#

#0 + 4 < -color(red)(1x) + 3x + 0#

#4 < (-color(red)(1) + 3)x#

#4 < 2x#

Now, divide each side of the inequality by #color(red)(2)# to solve for #x# while keeping the inequality balanced:

#4/color(red)(2) < (2x)/color(red)(2)#

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

#2 < x#

Or, we can reverse or "flip" the inequality to state the solution in terms of #x#:

#x > 2#