How do you solve #6+ 8x < 2( - 3+ 4x )#?

1 Answer
Oct 16, 2016

#x in O/#

Explanation:

Your goal here is to isolate #x# on one side of the inequality.

#6 + 8x < 2 * (-3 + 4x)#

Expand the parenthesis to get

#6 + 8x < 2 * (-3) + 2 * 4x#

#6 + 8x < - 6 + 8x#

Now, notice what happens when you subtract #8x# from both sides of the inequality

#6 + color(red)(cancel(color(black)(8x))) - color(red)(cancel(color(black)(8x))) < -6 + color(red)(cancel(color(black)(8x))) - color(red)(cancel(color(black)(8x)))#

#6 < -6#

When is #6# smaller than #-6#?

Never, which implies that this inequality has no real solutions. In other words, regardless of the value of #x# you plug in, the left side of the original inequality will never be smaller than the right side.

#6 + 8x color(red)(cancel(color(black)(<))) 2 * (-3 + 4x), " "(AA) x in RR#

Therefore, you can say that #x in O/#.