# How do you solve and write the following in interval notation: (x)/(x+1) ≤ (-4)/(3(x-4))?

Dec 10, 2017

See below

#### Explanation:

x/(x+1)≤−4/(3(x−4))

x/(x+1)+4/(3(x−4))≤0

(3x(x-4)+4(x+1))/(3(x+1)(x−4))≤0

(3x^2-8x+4)/(3(x+1)(x−4))≤0

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Let: a=3; b=-8; c=4

${x}_{1 , 2} = \frac{8 \pm \sqrt{{8}^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{8 \pm \sqrt{16}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{8 \pm 4}{2 \cdot 3}$

${x}_{1} = \frac{2}{3}$

${x}_{2} = 2$

((x-2/3)(x-2))/(3(x+1)(x−4))≤0

We need to find when is function $\le 0$
function is negative for:
$x \in \left(- 1 , \frac{2}{3}\right] \bigcup \left[2 , 4\right)$