anonymous 5 years ago how to solve x from x/2000 < 100x/(100x+1600)??

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

${x \over 2000 } < {100x \over 100x+1600}$ $x < {200000x \over 100x + 1600}$

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

now we are ready to solve this

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

so one part of the solution is x < -16

18. anonymous

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

19. anonymous

thus the full answer would be x < -16 or 0<x<1984

20. anonymous

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

21. anonymous

22. anonymous

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

23. anonymous

did it help at all ?

24. anonymous

hahahah u r using what anwar taught u xD

25. anonymous

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

26. anonymous

yes! thank you very much for your help :)