Simplify fully:?

#(9x^2-1)/(3x^2+2x-1)##div (3x+1)/(x-2)#

2 Answers
Jan 16, 2018

#(x-2)/(x+1)# when #x!=+-1/3#and#x!=-1#

Explanation:

First, remember that:
#(a/b)/(c/d)=a/b*d/c#
Therefore, #((9x^2-1)/(3x^2+2x-1))/((3x+1)/(x-2))=(9x^2-1)/(3x^2+2x-1)*(x-2)/(3x+1)#
Let's factor the denominator and the numerator of #(9x^2-1)/(3x^2+2x-1)#

#9x^2-1=(3x+1)(3x-1)#

We use the quadratic formula #(-b+-sqrt (b^2-4(a)(c)))/(2(a))#
#(-b+-sqrt (b^2-4(a)(c)))/(2(a))=x#
#(-2+-sqrt (2^2-4(3)(-1)))/(2(3))=x#
#(-2+-sqrt 16)/6=x#
#(-2+-4)/6=x#
#-1=x=1/3#
#3x^2+2x-1=3(x+1)(x-1/3)#

So we now have: #((3x+1)(3x-1))/(3(x+1)(x-1/3))*(x-2)/(3x+1)#

Now, remember that: #(ab)/(cd)*(ed)/(fg)=(ab)/(c canceld)*(ecanceld)/(fg)#

Therefore, we now have:

#((3x-1)(x-2))/(3(x+1)(x-1/3))=>((3x-1)(x-2))/((x+1)(3x-1))#

We see that both the denominator and the numerator share #3x-1# in common.

#(cancel(3x-1)(x-2))/((x+1)cancel(3x-1))#
#(x-2)/(x+1)# This is our answer!

Remember, however, that our original expression is undefined when
#x# is #+-1/3# or #-1#

Jan 17, 2018

#(9x^2-1)/(3x^2+2x-1) -: (3x+1)/(x-2) = (x-2)/(x+1) = 1-3/(x+1)#

with exclusion #x != +-1/3#

Explanation:

#(9x^2-1)/(3x^2+2x-1) -: (3x+1)/(x-2)#

#=(9x^2-1)/(3x^2+2x-1) * (x-2)/(3x+1)#

#=(color(red)(cancel(color(black)((3x-1))))color(blue)(cancel(color(black)((3x+1)))))/(color(red)(cancel(color(black)((3x-1))))(x+1)) * (x-2)/color(blue)(cancel(color(black)((3x+1))))#

#=(x-2)/(x+1)#

#=(x+1-3)/(x+1)#

#=1-3/(x+1)#

with exclusions #x != +-1/3#