How do you simplify #(ax - ay - 2by + 2bx)/(ax - 2bx + 2by + ay)#?

Question

Eva Adamson · Accepted Answer

#=>(x-y)/(((a-2b)/(a+2b))x+y)#

This cannot, in my opinion, be overly simplified:

#=>(ax - ay - 2by + 2bx)/(ax - 2bx + 2by + ay)#

I experimented by combining coefficients as follows:

#=>((a+2b)x-(a+2b)y)/((a-2b)x+(a+2b)y)#

I then noticed that the same coefficient was present in three of the four terms, so I divided by it:

#=>(x-y)/(((a-2b)/(a+2b))x+y)#

I guess this is a little better than the original, because whatever #a# and #b# happen to be, they only influence the one term in the denominator and it's easy to see how with this form.

Isaiah Anthony · Answer

undefined To simplify the expression (ax - ay - 2by + 2bx)/(ax - 2bx + 2by + ay), we can factor out common terms from both the numerator and the denominator.

First, let's factor out an 'a' from the numerator and the denominator:
(ax - ay - 2by + 2bx)/(ax - 2bx + 2by + ay) = (a(x - y) - 2b(y - x))/(a(x - y) - 2b(x - y))

Next, we can simplify further by canceling out the common factor of (x - y) from both the numerator and the denominator:
(a(x - y) - 2b(y - x))/(a(x - y) - 2b(x - y)) = (a - 2b)/(a - 2b)

Therefore, the simplified expression is (a - 2b)/(a - 2b).