# Thread: Inequality and Interval Notation?

1. ## Inequality and Interval Notation?

I have to solve the following inequality, answer written in interval notation

x/(x+3)<_ (-8)/(7(x-6))
BTW <_ is supposed to mean less than or equal to, I was unsure how to insert that.
I get this answer every time, which apparently is wrong:
7x^2-42x<_-8x -24
7x^2-34x+24<_0
(7x-6)(x-4)<_0
[6/7,4]

2. Originally Posted by katekate
I have to solve the following inequality, answer written in interval notation
x/(x+3)<_ (-8)/(7(x-6))
This is an equivalent problem.
$\displaystyle \frac{{(7x - 6)(x - 4)}}{{7(x + 3)(x - 6)}} \leqslant 0$.

3. I don't mean to sound dense but that really did not help me, I don't understand what you mean.

4. Originally Posted by katekate
I don't mean to sound dense but that really did not help me, I don't understand what you mean.
$\displaystyle \frac{x} {{x + 3}} \leqslant - \frac{8} {{7(x -6)}}\, \Rightarrow \,\frac{x} {{x + 3}} + \frac{8} {{7(x -6)}} \leqslant 0\, \Rightarrow \,\frac{{(7x - 6)(x - 4)}} {{7(x + 3)(x -6)}} \leqslant 0$

5. The equation was 7(x-6) in the second denominator, not +...not sure if this makes a huge difference but either way i don't understand how you went from the second step to the third.

6. I corrected the typo.

7. Okay, I've got it now. Thank you very much!