- Thu Aug 23, 2012 3:48 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solving Rational Equations (Imaginary Answer?)
- Replies:
**4** - Views:
**4088**

Oh yeah, I didn't see the (x+2) that would cancel on both sides...thanks for the help!

- Mon Aug 20, 2012 5:53 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solving Rational Equations (Imaginary Answer?)
- Replies:
**4** - Views:
**4088**

Ok these were my steps... (im going to explain in words instead of pictures because it took me forever to post that last message) 1. factored the 4-x^2 into -(x+2)(x-2). 2. then i multiplied top and bottom of x/(x-2) by -(x+2) so now it also has the -(x+2)(x-2) denominator. 3. then i combined the fr...

- Sun Aug 19, 2012 4:36 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solving Rational Equations (Imaginary Answer?)
- Replies:
**4** - Views:
**4088**

Problem: \frac{x}{x\, -\, 2}\, +\, \frac{2x}{4\, -\, x^2}\, =\, \frac{5}{x\, +\, 2} I can usually solve these kinds of problems pretty easily, but I got stuck with this one and when I checked the answer it was imaginary... I made a common denominator of -(x-2)(x+2) on the left side and then combined...

- Sun Aug 19, 2012 3:48 pm
- Forum: Trigonometry
- Topic: Solving Trig Inequalities...Need help.
- Replies:
**2** - Views:
**7310**

Thanks for the help, I think I can do it now. By the other interval you mean when both expressions are negative, right?

- Wed Aug 08, 2012 6:06 pm
- Forum: Trigonometry
- Topic: Solving Trig Inequalities...Need help.
- Replies:
**2** - Views:
**7310**

Hello everyone, I looked everywhere online and nobody properly explains how to solve trig inequalities, and i'm still stuck. This is the problem... 2sin 2 x > sinx , 0 < x < 2pi I got to this point... sinx(2sinx - 1) > 0 I then split the two expressions and solved, which gave me solutions of 0, pi, ...