How do you solve #-2/3x-4 < -8x-2#?

1 Answer
Apr 29, 2018

Answer:

#x < 3/11#

Explanation:

#-2/3x - 4 < -8x - 2#

First, add #4# to both sides of the inequality:
#-2/3x - 4 quadcolor(red)(+quad4) < -8x - 2 quadcolor(red)(+quad4)#

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

Add #8x# to both sides of the inequality:
#-2/3x quadcolor(red)(+quad8x) < -8x + 2 quadcolor(red)(+quad8x)#

#7 1/3x < 2#

#22/3x < 2#

Multiply both sides by #3#:
#22/3x color(red)(*quad3) < 2 color(red)(*quad3)#

#22/cancel(3)x cancel(color(red)(*quad3)) < 6#

#22x < 6#

#x < 6/22#

Divide numerator and denominator by #2#:
#x < 6/22 color(red)(-: 2/2)#

#x < 3/11#

Hope this helps!