John J.

asked • 11/20/17# Can you help me with my Precalculus?

the question I’m stuck on is “ solve 2/-6x - 3x/x-4 is greater than or equal to 0”

sorry I had to write it out different, there aren’t many options lol. If you can help me out, that would be great :)

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## 1 Expert Answer

Andy C. answered • 11/24/17

Math/Physics Tutor

Looks like the parenthesis should go like this:

-1/(3x) - 3x/(x-4) = 0 <--- otherwise the x will cancel in the second term, which is highly unusual

We need to find the zeros anyway, so that we can get the intervals for x that need to be tested.

The common denominator is (3x)(x-4).

-1(x-4) - 3x(3x) = 0

4 - x - 9x^2 = 0

9x^2 + x - 4 = 0

quadratic formula:

(-1 +or- sqrt( 1^2 - 4(9)(-4))/(2*9)

(-1 +or- sqrt( 1 + 144))/18

(-1 +or- sqrt(145))/18 <--- two zeros, both are irrational

There are vertical asymtopes at x=0 and x=4

The intervals of interest are

x< (-1 - sqrt(145))/18

**(-1 - sqrt(145))/18 < x < 0**

0 < x < (-1+sqrt(145))/18

**(-1 +sqrt(145))/18 <**

**x < 4**

x>4

Examining the graph, the intervals in bold face are the solutions

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Michael J.

11/20/17