## anonymous 5 years ago what is the solution in interval notation? -5/3x le -10

1. anonymous

-5/3x≤-10 multiply each side by -3/5 (don't forget to switch the sign!) x≥6 in interval notation [6, infty)

2. anonymous

its -5 over 3 x $\le$ -10

3. anonymous

how do i write it in interval notation?

4. anonymous

ooooh got it.

5. anonymous

with 6,infinitie)

6. anonymous

$[6, \infty)$

7. anonymous

thank u

8. anonymous

-5/3x ≤ -10 3x ≤ 1/2 x ≤ 1/6 which is to say that x is an element of (epsilon) (0,1/6] (parentheses because you can't have x=0, and a bracket because the expression equals -10 if x = 1/6). TBates, if you plug in 6, or anything greater, into the original inequality you don't even come close to approaching -10. In your method, you forgot that x is in the denominator -- not the numerator. If you graph -5/3x, you have a rational expression with asymptotes. You can use that to find the values at which the function is less than -10: in this circumstance, any value between 0 and 1/6 gives -10 or less. Anything less than 0 gives a value higher than -10, and it certainly cannot equal zero.

9. anonymous

but that answer is still correct...

10. anonymous

Quantum you forgot that the problem had 3 in the denominator, the answer I gave previously is correct.

11. anonymous

How it was written the x is not in parenthesis and therefor not a part of the denominator.

12. anonymous

So it was 5/3 * x? If so, then yeah, you're right. I thought it was 5/(3x).

13. anonymous

sorry, that was my fault. it was hard to write it on here and show the 3 was not part of the x