How do you solve #4y - 3< 2y + 2#b?

1 Answer
smendyka
Sep 3, 2017

See a solution process below:

Explanation:

If You Are Solving For y:

First, add #color(red)(3)# and subtract #color(red)(2y)# from each side of the inequality to isolate the #y# term while keeping the inequality balanced:

#-color(red)(2y) + 4y - 3 + color(red)(3) < -color(red)(2y) + 2y + 2b + color(red)(3)#

#(-color(red)(2) + 4)y - 0 < 0 + 2b + 3#

#2y < 2b + 3#

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

#(2y)/color(red)(2) < (2b + 3)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) < (2b + 3)/2#

#y < (2b + 3)/2#

Or

#y < (2b)/2 + 3/2#

#y < (color(red)(cancel(color(black)(2)))b)/color(red)(cancel(color(black)(2))) + 3/2#

#y < b + 3/2#

If You Are Solving For b:

First, subtract #color(red)(2y)# from each side of the inequality to isolate the #b# term while keeping the inequality balanced:

#-color(red)(2y) + 4y - 3 < -color(red)(2y) + 2y + 2b#

#(-color(red)(2) + 4)y - 3 < 0 + 2b#

#2y - 3 < 2b#

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

#(2y - 3)/color(red)(2) < (2b)/color(red)(2)#

#(2y - 3)/2 < (color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2))#

#(2y - 3)/2 < b#

If we want to state the solution in terms of #b# we can reverse or "flip" the entire inequality:

#b > (2y - 3)/2#

Or

#b > (2y)/2 - 3/2#

#b > (color(red)(cancel(color(black)(2)))y)/color(red)(cancel(color(black)(2))) - 3/2#

#b > y - 3/2#

Related questions
Impact of this question
502 views around the world
You can reuse this answer
Creative Commons License