How do you combine #a/(a-3)-5/(a+6)#?

To subtract these two fractions, as with any fractions, they need to be over common denominators. In this case the common denominator will be #(a - 3)(a + 6)#.

Therefore, first we need to multiply each fraction by the correct form of #1# to put each fraction over the common denominator:

#(color(red)(((a + 6))/((a + 6))) xx a/((a - 3))) - (color(blue)(((a - 3))/((a - 3))) xx 5/((a + 6)))#

#((a + 6)a)/((a + 6)(a - 3)) - ((a - 3)5)/((a - 3)(a + 6)#

We can now expand the terms within parenthesis in each of the numerators:

#(a^2 + 6a)/((a + 6)(a - 3)) - (5a - 15)/((a - 3)(a + 6)#

We can now combine the fractions by subtracting the numerators and keeping the common denominator:

#((a^2 + 6a) - (5a - 15))/((a - 3)(a + 6)#

Now we can expand the terms within the parenthesis ensuring we keep the signs of the terms correct:

#(a^2 + 6a - 5a + 15)/((a - 3)(a + 6)#

And then we can combine like terms in the numerator:

#(a^2 + (6 - 5)a + 15)/((a - 3)(a + 6)#

#(a^2 + a + 15)/((a - 3)(a + 6)#

Or, is we want to expand the term in the denominator by cross multiplying we get:

#(a^2 + a + 15)/(a^2 + 6a - 3a - 18)#

#(a^2 + a + 15)/(a^2 + (6 - 3)a - 18)#

#(a^2 + a + 15)/(a^2 + 3a - 18)#