Expert: Paul Klarreich - 5/16/2007

Question
hi i have this question i am attempting. (i)Find a,b E R such that [[(3x)/(x+1)] - 2]=(ax+b)/(x+1) where x not=-1 AND (ii) HENCE, find range of values of x E R for which
[(3x)/(x+1)]>2
this is what i tried:  for (i) i got [(x-2)/(x+1)]= (ax+b)/(x+1)therefore a=1 and b=-2 BUT for (ii) they use the word HENCE...  and i used normal Arithmetic to end with the result x>2 but how do i arrive at this answer from part (i) (how do i use it)? thanks a million in advance  

Answer
Questioner:   jon
Category:  Calculus
 
Subject:  math problem
Question:  hi i have this question i am attempting. (i)Find a,b E R such that [[(3x)/(x+1)] - 2]=(ax+b)/(x+1) where x not=-1 AND (ii) HENCE, find range of values of x E R for which [(3x)/(x+1)]>2

this is what i tried:  for (i) i got [(x-2)/(x+1)]= (ax+b)/(x+1)therefore a=1 and b=-2

BUT for (ii) they use the word HENCE...  

>> You let them psych you with that word.  Shame!

and i used normal Arithmetic to end with the result x>2 but how do i arrive at this answer from part (i) (how do i use it)? thanks a million in advance
.............................
Hi, Jon,

This is your form:

 3x         ax + b
------ - 2 = --------
x + 1         x + 1
 
3x  - 2(x + 1)    ax + b
--------------  = --------
  x + 1           x + 1

3x  - 2x - 2       ax + b
--------------  = --------
  x + 1           x + 1

x - 2      ax + b
-------  = -------
x + 1       x + 1

a = 1,  b = -2

Yes, that confirms what you did.

Now about this inequality:

[(3x)/(x+1)]>2, or:

 3x
------ > 2
x + 1
That is equivalent to:

 3x
------ - 2 > 0
x + 1

and the left side is the same as this, after the first part of the algebra:

x - 2
----- > 0
x + 1

Now how about this inequality?  It says that some fraction (x-2 over x+1) is positive.  When is a fraction positive?  Answer: When the top and bottom have the same sign.  That means:

x - 2 > 0  and  x + 1 > 0
x > 2  and  x > - 1

The intersection of those is  x > 2.  So far so good.  This is what you got.

BUT IT COULD ALSO BE:

x - 2 < 0  and  x + 1 < 0
x < 2  and  x < -1

The intersection of those is  x < -1.  So the answer to the second question is: