How do you simplify #((x-5)/(x+3)) /(( x-2)/(x+3))#?

1 Answer
Apr 19, 2016

#=(x-5)/(x-2)#

Explanation:

Right now, that equation can look too tall to deal with, so lets just put it into two fractions:
#((x-5)/(x+3))/((x-2)/(x+3))=(x-5)/(x+3) divide (x-2)/(x+3)#
Because we know that to divide by a fraction you just multiply by its reciprocal (its flipped version), we can simplify the whole thing:
#=(x-5)/(x+3) * (x+3)/(x-2)#
As you can see, we can cancel out the #x+3#, and write it as one fraction:
#=(x-5)/(x-2)#

Another way to see the question is like this:
#((x-5)/(x+3))/((x-2)/(x+3))=((x-5)/(x+3))/((x-2)/(x+3))*(x+3)/(x+3)#
#=(x-5)/(x-2)#
where you just multiply the top and bottom by the same thing to remove the fractions at the top and bottom