How do you solve # 4/3r - 3 < r + 2/3 - 1/3r#?

1 Answer
May 3, 2017

Answer:

See the solution process below:

Explanation:

First, multiply each side of the inequality by #color(red)(3)# to eliminate the fractions while keeping the inequalitybalanced:

#color(red)(3)(4/3r - 3) < color(red)(3)(r + 2/3 - 1/3r)#

#(color(red)(3) * 4/3r) - (color(red)(3) * 3) < (color(red)(3) * r) + (color(red)(3) * 2/3) - (color(red)(3) * 1/3r)#

#(cancel(color(red)(3)) * 4/color(red)(cancel(color(black)(3)))r) - 9 < 3r + (cancel(color(red)(3)) * 2/color(red)(cancel(color(black)(3)))) - (cancel(color(red)(3)) * 1/color(red)(cancel(color(black)(3)))r)#

#4r - 9 < 3r+ 2 - 1r#

Next, group and combine like terms on the left side of the inequality:

#4r - 9 < 3r - 1r + 2#

#4r - 9 < (3 - 1)r + 2#

#4r - 9 < 2r + 2#

Then, add #color(red)(9)# and subtract #color(blue)(2r)# from each side of the inequality to isolate the #r# term while keeping the inequality balanced:

#-color(blue)(2r) + 4r - 9 + color(red)(9) < -color(blue)(2r) + 2r + 2 + color(red)(9)#

#(-color(blue)(2) + 4)r - 0 < 0 + 11#

#2r < 11#

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

#(2r)/color(red)(2) < 11/color(red)(2)#

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

#r < 11/2#