How do you solve #3(b -5)<-2b#?

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Answer 1 · 2017-10-11T21:04:22

See a solution process below:

First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(3)(b - 5) < -2b#

#(color(red)(3) xx b) - (color(red)(3) xx 5) < -2b#

#3b - 15 < -2b#

Next, add #color(red)(15)# and #color(blue)(2b)# to each side of the inequality to isolate the #b# term while the keeping the inequality balanced:

#color(blue)(2b) + 3b - 15 + color(red)(15) < color(blue)(2b) - 2b + color(red)(15)#

#(color(blue)(2) + 3)b - 0 < 0 + color(red)(15)#

#5b < 15#

Now, divide each side of the inequality by #color(red)(5)# to solve for #b# while keeping the inequality balanced:

#(5b)/color(red)(5) < 15/color(red)(5)#

#(color(red)(cancel(color(black)(5)))b)/cancel(color(red)(5)) < 3#

#b < 3#