How do you solve #(b+3)/(5-2b)<=4# using a sign chart?

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Answer 1 · 2017-09-05T12:16:39

Solution: # b <=17/9 and b>2.5 or (-oo,17/9]uu (2.5 , oo)#

# (b+3)/(5-2b) <= 4 or (b+3)/(5-2b)-4 <=0# or

# ((b+3)-4(5-2b))/(5-2b) <=0 or (9b-17)/(5-2b) <=0#

Critical points are # 9b=17 or b=17/9 and 2b=5 or b=2.5#

#b !=2.5 # as denominator should not be zero.

Sign chart:

# b< 17/9#, sign of #(9b-17)/(5-2b) # is #(-)/(+)=(-) , < 0#

# 17/9 < b < 2.5#, sign of #(9b-17)/(5-2b) # is #(+)/(+)=(+) , > 0#

# b > 2.5#, sign of #(9b-17)/(5-2b) # is #(+)/(-)=(-) , < 0#

Solution: # b <=17/9 and b>2.5 or (-oo,17/9]uu (2.5 , oo)#

graph{(9x-17)/(5-2x) [-28.48, 28.47, -14.24, 14.25]} [Ans]