How do you simplify #(a^2-5a)/(3a-18)-(7a-36)/(3a-18)#?

1 Answer
Mar 15, 2017

#(a^2-5a)/(3a-18)-(7a-36)/(3a-18) = a/3-2#

with exclusion #a!=6#

Explanation:

Since the denominators are identical, we can start by simply subtracting the numerators:

#(a^2-5a)/(3a-18)-(7a-36)/(3a-18) = (a^2-5a-7a+36)/(3a-18)#

#color(white)((a^2-5a)/(3a-18)-(7a-36)/(3a-18)) = (a^2-12a+36)/(3a-18)#

#color(white)((a^2-5a)/(3a-18)-(7a-36)/(3a-18)) = ((a-6)(color(red)(cancel(color(black)(a-6)))))/(3(color(red)(cancel(color(black)(a-6)))))#

#color(white)((a^2-5a)/(3a-18)-(7a-36)/(3a-18)) = (a-6)/3#

#color(white)((a^2-5a)/(3a-18)-(7a-36)/(3a-18)) = a/3-2#

with exclusion #a != 6#

Note that if #a=6#, then the original expression is undefined, but the simplified one is defined. So we need to specify that #a=6# is excluded.