To solve this inequality, we need to find the critical points where the expression is equal to zero and then test the intervals between these critical points to determine where the expression is greater than zero.
First, let's find the critical points by setting the expression equal to zero:
(x+5)(x-7)/(3x-1) = 0
This equation is true when the numerator is equal to zero:
x+5 = 0 --> x = -5 x-7 = 0 --> x = 7
Now, we have critical points at x = -5 and x = 7.
Next, we test the intervals (-∞, -5), (-5, 7), and (7, ∞) one by one to see where the expression is greater than zero.
For the interval (-∞, -5): Choose x = -6, substitute it into the expression: (-6+5)(-6-7)/(3(-6)-1) = (-1)(-13)/(-18-1) = 13/19 > 0
For the interval (-5, 7): Choose x = 0, substitute it into the expression: (0+5)(0-7)/(3(0)-1) = (5)(-7)/(-1) = -35 < 0
For the interval (7, ∞): Choose x = 8, substitute it into the expression: (8+5)(8-7)/(3(8)-1) = (13)(1)/(23) = 13/23 > 0
Therefore, the solution to the inequality (x+5)(x-7)/(3x-1) > 0 is x ∈ (-∞, -5) U (7, ∞).
To solve this inequality, we need to find the critical points where the expression is equal to zero and then test the intervals between these critical points to determine where the expression is greater than zero.
First, let's find the critical points by setting the expression equal to zero:
(x+5)(x-7)/(3x-1) = 0
This equation is true when the numerator is equal to zero:
x+5 = 0 --> x = -5
x-7 = 0 --> x = 7
Now, we have critical points at x = -5 and x = 7.
Next, we test the intervals (-∞, -5), (-5, 7), and (7, ∞) one by one to see where the expression is greater than zero.
For the interval (-∞, -5):
Choose x = -6, substitute it into the expression:
(-6+5)(-6-7)/(3(-6)-1) = (-1)(-13)/(-18-1) = 13/19 > 0
For the interval (-5, 7):
Choose x = 0, substitute it into the expression:
(0+5)(0-7)/(3(0)-1) = (5)(-7)/(-1) = -35 < 0
For the interval (7, ∞):
Choose x = 8, substitute it into the expression:
(8+5)(8-7)/(3(8)-1) = (13)(1)/(23) = 13/23 > 0
Therefore, the solution to the inequality (x+5)(x-7)/(3x-1) > 0 is x ∈ (-∞, -5) U (7, ∞).