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]

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