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

1 Answer
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)=(-b+-sqrt(b^2-4ac))/(2a)

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

x_(1,2)=(8+-sqrt(8^2-4*3*4))/(2*3)

x_(1,2)=(8+-sqrt(16))/(2*3)

x_(1,2)=(8+-4)/(2*3)

x_1=2/3

x_2=2

((x-2/3)(x-2))/(3(x+1)(x−4))≤0

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We need to find when is function <=0
function is negative for:
x in (-1,2/3]uuu[2,4)