Simplifying . adding rational expressions answer asap with steps please

Question

(n+3)/(n²-5n+6)  +  (6)/(n²-7n+12)

Answer 1

(n+3)/(n²-5n+6) + (6)/(n²-7n+12)

 

Factor the denominators:  (n+3)/((n-3)(n-2)) + (6)/((n-3)(n-4))

 

Note that the LCD is (n-3)(n-2)(n-4).

 

Now, multiply the first fraction by the missing factor:  (n-4) and the second fraction by its missing factor: (n - 2)

 

We now have: ((n+3)(n-4))/((n-3)(n-2)(n-4)) + ((6(n-2)/((n-3)(n-4)(n-2)).

 

Putting these together: (n² - n - 12 + 6n - 12)/((n-3)(n-4)(n-2)

 

... or (n² + 5n - 24)/((n-3)(n-4)(n-2))

 

Now, factor the numerator:  ((n-3)(n+8))/((n-3)(n-4)(n-2))

 

Since (n-3)/(n-3) equals 1, the answer becomes:  (n+8)/((n-4)(n-2)).

 

[Note:  I didn't use the common word "cancel" because every year, at least one student believes that that makes the term disappear instead of becoming 1.  i.e. (25x² - 15x + 5)/5 becoming 5x² - 3x, which I see at least once per class.]

Answer 2

Since I read this as an addition of fractions problem, I offer this solution. (Josh: Thanks for using the parentheses to clarify your problem!)

(n+3)/(n²-5n+6) + (6)/(n²-7n+12) needs the denominators factored into

(n+3)/(n-3)(n-2) + (6)/(n-4)(n-3).

Then use a common denominator of (n-4)(n-3)(n-2) to rewrite the fractions.

(n+3)(n-4)/(n-4)(n-3)(n-2) + (6)(n-2)/(n-4)(n-3)(n-2).

First multiply the terms in each numerator to get

n²-n-12 and 6n-12 then add them to get n²+5n-24 which factors to become (n-3)(n+8). This is the numerator.

Now the sum is (n-3)(n+8)/(n-4)(n-3)(n-2) with a common factor of (n-3), it reduces to (n+8)/(n-4)(n-2) as your final answer.