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

Sep 5, 2017

Solution: $b \le \frac{17}{9} \mathmr{and} b > 2.5 \mathmr{and} \left(- \infty , \frac{17}{9}\right] \cup \left(2.5 , \infty\right)$

#### Explanation:

$\frac{b + 3}{5 - 2 b} \le 4 \mathmr{and} \frac{b + 3}{5 - 2 b} - 4 \le 0$ or

$\frac{\left(b + 3\right) - 4 \left(5 - 2 b\right)}{5 - 2 b} \le 0 \mathmr{and} \frac{9 b - 17}{5 - 2 b} \le 0$

Critical points are $9 b = 17 \mathmr{and} b = \frac{17}{9} \mathmr{and} 2 b = 5 \mathmr{and} b = 2.5$

$b \ne 2.5$ as denominator should not be zero.

Sign chart:

$b < \frac{17}{9}$, sign of $\frac{9 b - 17}{5 - 2 b}$ is $\frac{-}{+} = \left(-\right) , < 0$

$\frac{17}{9} < b < 2.5$, sign of $\frac{9 b - 17}{5 - 2 b}$ is $\frac{+}{+} = \left(+\right) , > 0$

$b > 2.5$, sign of $\frac{9 b - 17}{5 - 2 b}$ is $\frac{+}{-} = \left(-\right) , < 0$

Solution: $b \le \frac{17}{9} \mathmr{and} b > 2.5 \mathmr{and} \left(- \infty , \frac{17}{9}\right] \cup \left(2.5 , \infty\right)$

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