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

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) (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:

must be positive (or zero) because sqaring a Real number will never result in a negative.
Since 3 and are both positive (or zero) must also be positive (or zero).
Since must be positive (or zero) 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":

Solving this inequality we subtract 2 from both sides

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.)

So x must be less than . In interval notation this is (-infinity, 1/3).
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