How do you simplify #(2t)/(s^2 - st) + (s+t)/(t^2 - s^2)#?

1 Answer
Oct 20, 2015

#(s - 2t)/(s(t-s))#

Explanation:

Your starting expression looks like this

#(2t)/(s^2 - st) + (s + t)/(t^2 - s^2)#

Notice that you can write

#s^2 - st = - (st - s^2)#

The expression becomes

#-(2t)/(st - s^2) + (s + t)/(t^2 - s^2)#

Now focus on factoring the two denominators. For #st - s^2#, you can use #s# as a common factor to write

#st - s^2 = s * (t - s)#

Notice that the second denominator is a difference of squares, which means that you can factor it using

#color(blue)(a^2 - b^2 = (a-b)(a+b))#

In your case, you would have

#t^2 - s^2 = (t-s)(t+s)#

The expression can thus be written as

#-(2t)/(s(t-s)) + color(red)(cancel(color(black)((t+s))))/((t-s) color(red)(cancel(color(black)((t+s)))))#

#-(2t)/(s(t-s)) + 1/(t-s)#

Multiply the second fraction by #1 = s/s# to get

#-(2t)/(s(t-s)) + s/(s(t-s)) = color(green)((s - 2t)/(s(t-s))#