Question

WILL GIVE A BRAINLESTTTT What is the solution of 3x+8/x-4 >= 0

Answer 1

Answer:

The inequality is given to be :

$\frac{3x+8}{x-4}\geq 0$

The inequality will be greater than or equal to 0 if and only if both the numerator and denominator of the left hand side will have same sign either both positive or both negative.

CASE 1 : Both positive

3x + 8 ≥ 0

⇒ 3x ≥ -8

$x\geq \frac{-8}{3}$

Also, x - 4 ≥ 0

⇒ x ≥ 4

Now, Taking common points of both the values of x

⇒ x ∈ [4, ∞)

CASE 2 :  Both are negative

3x + 8 ≤ 0

⇒ 3x ≤ -8

$x\leq \frac{-8}{3}$

Also, x - 4 ≤ 0

⇒ x ≤ 4

So, Taking common points of both the values of x we have,

$x=(-\infty,-\frac{8}{3}]$

So, The solution of the equation will be the union of both the two solutions

So, Solution is given by :

$x=(-\infty,-\frac{8}{3}]\:U\:[4,\infty)$

Answer 2

Solution: (-Infinite, -8/3] U (4, Infinite)

Using that a fraction is greater than or equal to zero when the numerator and denominator have the same sign:
a/b>=0. Then we have two cases:
Case 1) If the numerator is positive, the denominator must be positive too (at the same time):
if a>=0 ∩ b>0

Or (U)

Case 2) If the numerator is negative, the denominator must be negative too (at the same time):
if a<=0 ∩ b<0

In this case a=3x+8 and b=x-4, then:

Case 1):
if 3x+8>=0 ∩ x-4>0
Solving for x:
3x+8-8>=0-8 ∩ x-4+4>0+4
3x>=-8 ∩ x>4
3x/3>=-8/3 ∩ x>4
x>=-8/3 ∩ x>4
Solution Case 1: x>4 = (4, Infinite)

Case 2):
if 3x+8<=0 ∩ x-4<0
Solving for x:
3x+8-8<=0-8 ∩ x-4+4<0+4
3x<=-8 ∩ x<4
3x/3<=-8/3 ∩ x<4
x<=-8/3 ∩ x<4
Solution Case 2: x<=-8/3 = (-Infinite, -8/3]

Solution= Solution Case 1 U Solution Case 2
Solution = (4, Infinite) U (-Infinite, -8/3]
Solution: (-Infinite, -8/3] U (4, Infinite)