SOLUTION: Solve the inequality {(-6x+2)/(3x^2+7)}>0 possible answers: (-1/3,infinite) (-infinite,1/3) (-infinite,0) (-infinite,-3)

Algebra ->  Equations -> SOLUTION: Solve the inequality {(-6x+2)/(3x^2+7)}>0 possible answers: (-1/3,infinite) (-infinite,1/3) (-infinite,0) (-infinite,-3)      Log On


   



Question 201394: Solve the inequality
{(-6x+2)/(3x^2+7)}>0
possible answers:
(-1/3,infinite)
(-infinite,1/3)
(-infinite,0)
(-infinite,-3)

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Your problem says that a fraction is greater than zero. In other words, the fraction is positive. And if we think about how fractions can be positive (and how they can be negative) we should be able to determine that there are two possibilities: The numerator and denominator are both positive or they are both negative.
If we look at the denominator we can tell the following:
  1. x%5E2 must be positive (or zero) because sqaring a Real number will never result in a negative.
  2. Since 3 and x%5E2 are both positive (or zero) 3x must also be positive (or zero).
  3. Since 3x must be positive (or zero) 3x%2B7 must be positive.
.Since the denominator of our fraction must be positive, the numerator must also be positive (in order for the fraction as a whole to be positive. So now we know that the numerator must be positive. In other "words":
-6x%2B2+%3E+0
Solving this inequality we subtract 2 from both sides
-6x+%3E+-2
Divide by -6. (Remember the special rule that applies when multiplying or dividing both sides of an inequality by a negative number! We must reverse the inequality.)
x+%3C+1%2F3
So x must be less than 1%2F3. In interval notation this is (-infinity, 1/3).