How do you solve #4 > 3x + 3 or 9 ≤ - 5x + 3#?

2 Answers
Jul 1, 2015

Answer:

Explanation:

(1) 4 > 3x + 3 --> 3x < 4 - 3 --> #x < 1/3#

(2) #9 <= - 5x + 3 -> - 5x >= 6# --> #x <= -6/5#

Jul 1, 2015

Answer:

#x<1/3#

Explanation:

Part 1: #4>3x+3#
#4>3x+3#
#color(white)("XXXX")#subtract 3 from both sides
#1>3x#
#color(white)("XXXX")#divide both sides by 3
#1/3>x#

Part 2: #9<=−5x+3#
#9<=-5x+3#
#color(white)("XXXX")#subtract 3 from both sides
#6<=-5x#
#color(white)("XXXX")#divide both sides by #(-5)# remembering to reverse the inequality orientation because of the negative division.
#-6/5>=x#

Part 3: combining with or
If it is only necessary for
#color(white)("XXXX")##x<1/3#
or
#color(white)("XXXX")##x<=-6/5#
then
it is only necessary for
#color(white)("XXXX")##x<1/3#
#color(white)("XXXX")##color(white)("XXXX")#since #(x<=-6/5)# is a subset of #(x<1/3)#)