## anonymous 4 years ago 3|x| / x-1 < 2

1. anonymous

2. Mertsj

b is easy. 3x-2<1 and 3x-2>-1 x<1 and x>1/3

3. anonymous

dn't get it..can you please attach?? :(

4. anonymous

its the first question on attachment..please have a look..thanks :)

5. Mertsj

|3x-2|<1 means 3x-2 is less than 1 unit from 0 which means 3x-2 <1 and 3x-2>-1 Just solve each part.

6. anonymous

3 |x| / ( x-1) < 2 this is the question... :'(

7. anonymous

humm, I have problem with a and c :(

8. Mertsj

I think just by inspection that x <1 is the solution to the first one.

9. anonymous

this question is really confusing :@..

10. Mertsj

The third one, let's just take the log of both sides and solve.

11. Mertsj

x>.79

12. anonymous

humm I am trying with log..

13. Mertsj

$(2x-1)\log_{10}2<(3x-2)\log_{10}3$

14. Mertsj

$(2x-1)(.301)<(3x-2)(.477)$

15. anonymous

are we allowed to use the log value...(if i'm not allowed to bring my calculator then how will i solve).. :(

16. Mertsj

The logs you need would have to be provided.

17. Mertsj

Are calculators forbidden?

18. anonymous

I'm not sure..bt, my friend told me.. :'(

19. anonymous

thanks man for your efforts... :)

20. Mertsj

yw

21. Mertsj

for the first one, I found these steps: 1. Move all terms to one side of the inequality sign so that one side is 0. 2. Replace the inequality sign with an equal sign and solve the equation. These solutions are critical values 3. find all values that result in division by 0. These are critical values 4. plot the critical values on a number line. 5. Test each interval defined by the critical values. If an interval satisfies the inequality then it is part of the solution.

22. Mertsj

$\frac{3|x|}{x-1}-2=0$

23. Mertsj

I found three critical numbers: -2, 2/5 and 1

24. Mertsj

And the inequality is true in all intervals except x>1 so just as I thought, the solution is x<1

25. anonymous

hummm...but, its too complicated....are we allowed to use equal sign for inequality? I'm not sure about this process...

26. Mertsj

It is only as an aid to find the critical numbers. It is not part of the solution.

27. anonymous

humm got it now..

28. Mertsj

ok

29. anonymous

thanks again.. :)

30. Mertsj

yw