how to solve x from x/2000 < 100x/(100x+1600)??

- anonymous

how to solve x from x/2000 < 100x/(100x+1600)??

- chestercat

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- yuki

these questions are more sensitive than we all think, so let's go through it carefully

- yuki

remember that when we multiply negative numbers in inequalities the < has to switch into > ?

- yuki

since you have a denominator that has a variable on the right side, (100x+1600) we are not really allowed to multiply it on both sides and cancel it because we don't know if the number is negative or positive, so this is the step that you have to take

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## More answers

- anonymous

i forgot to mention it. x is non-negative.

- yuki

\[{x \over 2000 } < {100x \over 100x+1600}\]
\[x < {200000x \over 100x + 1600}\]

- yuki

so far I multiplied positive numbers so it is ok,
now what you have to do is to gather all expressions on
one side and make it into one big fraction as follows

- yuki

\[x- {200000x \over 100x+1600} < 0\]
\[{x(100x+1600) \over 100x + 1600} -{200000x \over 100x + 1600}\] <0

- yuki

\[{100x^x+1600x - 200000x \over 100x+1600} <0\]

- yuki

\[{100x^2 - 198400x \over 100x+1600} < 0\]

- yuki

lets factor out the 100s (I should have done this first to make the calculations look better :\ )
\[ {x^2 - 984x \over x + 16} <0\]

- yuki

now we are ready to solve this

- yuki

the numerator factors into x (x-1984) and the denominator is x+16

- yuki

this is what you are going to do
check the intervals between the x's that makes the numerator = 0 and the denominator = 0

- yuki

we can easily see that the numerator = 0 when
x=0 or x = 1984 and the denominator =0 when x = -16
so the interval we check is
---(-16)----(0)------(1984)----
the left of -6, between -16 and 0, between 0 and 1984 and after 1984

- yuki

the idea is this , all we need to do is to check whether
\[x(x-1984) \over x+16\]
becomes negative or not,
so we will plug in numbers that falls into the above interval
and check for the sign

- yuki

for example less than -16
we can plug in -100 to see the sign of each factors
it becomes
\[(-)(-) \over(-)\]
which is negative

- yuki

so one part of the solution is x < -16

- yuki

now we check the rest
I will let you try doing the other parts, but between 0 and 1984
we can confirm that the rational expression will become negative
and the rest positive as such
if we plugged in x=1 into the rational expression, we get
\[(+)(-) \over (+)\]
which is negative

- yuki

thus the full answer would be
x < -16 or 0

- yuki

lets' summarize
1), gather all rational expressions on one side
2), make it into one big fraction and factor the top and bottom
3), find out the critical points (numerator = 0 and denominator = 0) draw them on a number line
4), you plug in any number that is between the critical points and check for the sign
5), the one you are looking for > or < will be the answer

- yuki

I just saw your reply lol
the answer will be 0 0

- yuki

but it's ok, we did not waste any time at all.
this is exactly how you can solve these inequalities.

- yuki

did it help at all ?

- anonymous

hahahah u r using what anwar taught u xD

- anonymous

and yes VERYYYY much im applying them to my problems like RITE now lol

- anonymous

yes! thank you very much for your help :)

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