3|x| / x-1

Question

3|x| / x-1 < 2

anonymous · Answer

attachment

mertsj · Answer

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

anonymous · Answer

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

anonymous · Answer

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

mertsj · Answer

|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.

anonymous · Answer

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

anonymous · Answer

humm,  I have problem with a and c :(

mertsj · Answer

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

anonymous · Answer

this question is really confusing :@..

mertsj · Answer

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

mertsj · Answer

x>.79

anonymous · Answer

humm I am trying with log..

mertsj · Answer

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

mertsj · Answer

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

anonymous · Answer

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

mertsj · Answer

The logs you need would have to be provided.

mertsj · Answer

Are calculators forbidden?

anonymous · Answer

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

anonymous · Answer

thanks man for your efforts... :)

mertsj · Answer

yw

mertsj · Answer

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.

mertsj · Answer

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

mertsj · Answer

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

mertsj · Answer

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

anonymous · Answer

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

mertsj · Answer

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

anonymous · Answer

humm got it now..

mertsj · Answer

ok

anonymous · Answer

thanks again.. :)

mertsj · Answer

yw