How do you solve and write the following in interval notation: $\frac{x}{x + 1} \leq \frac{- 4}{3 (x - 4)}$ $(x)/(x+1) ≤ (-4)/(3(x-4))$ ?

Question

2199 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-10T23:08:37

See below

$\frac{x}{x + 1} \leq - \frac{4}{3 (x - 4)}$ $x/(x+1)≤−4/(3(x−4))$

$\frac{x}{x + 1} + \frac{4}{3 (x - 4)} \leq 0$ $x/(x+1)+4/(3(x−4))≤0$

$\frac{3 x (x - 4) + 4 (x + 1)}{3 (x + 1) (x - 4)} \leq 0$ $(3x(x-4)+4(x+1))/(3(x+1)(x−4))≤0$

$\frac{3 x^{2} - 8 x + 4}{3 (x + 1) (x - 4)} \leq 0$ $(3x^2-8x+4)/(3(x+1)(x−4))≤0$

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x_{1, 2} = \frac{8 \pm \sqrt{8^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$ $x_(1,2)=(8+-sqrt(8^2-4*3*4))/(2*3)$

$x_{1, 2} = \frac{8 \pm \sqrt{16}}{2 \cdot 3}$ $x_(1,2)=(8+-sqrt(16))/(2*3)$

$x_{1, 2} = \frac{8 \pm 4}{2 \cdot 3}$ $x_(1,2)=(8+-4)/(2*3)$

$x_{1} = \frac{2}{3}$ $x_1=2/3$

$x_{2} = 2$ $x_2=2$

$\frac{(x - \frac{2}{3}) (x - 2)}{3 (x + 1) (x - 4)} \leq 0$ $((x-2/3)(x-2))/(3(x+1)(x−4))≤0$

enter image source here

We need to find when is function $\leq 0$ $<=0$
function is negative for:
$x \in (- 1, \frac{2}{3}] ⋃ [2, 4)$ $x in (-1,2/3]uuu[2,4)$

How do you solve and write the following in interval notation: (x)/(x+1) ≤ (-4)/(3(x-4))xx+1≤−43(x−4)(x)/(x+1) ≤ (-4)/(3(x-4))?