Question

what is the solution in interval notation? -5/3x le -10

Answer 1

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

Answer 2

its -5 over 3 x
\[\le\] -10

Answer 3

how do i write it in interval notation?

Answer 4

ooooh got it.

Answer 5

with 6,infinitie)

Answer 6

\[[6, \infty)\]

Answer 7

thank u

Answer 8

-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.

Answer 9

but that answer is still correct...

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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