• December 1st 2007, 12:55 PM
t-dot
I can't get the same answer as the book in this one:

$\frac{n^2+4n-3}{n^2-16}+\frac{4-3n}{3n-12}$

the answer i ended up with:

$\frac{3n+3}{3(n+4)(n-4)}$

$\frac{4n+7}{3(n+4)(n-4)}$
• December 1st 2007, 01:23 PM
Soroban
Hello, t-dot!

If you'd show your work, we could find your error . . .

Quote:

[tex] $\frac{n^2+4n-3}{n^2-16}+\frac{4-3n}{3n-12}$

We have: . $\frac{n^2 + 4n - 3}{(n+4)(n-4)} + \frac{4-3n}{3(n-4)}$

Get a common denominator: . ${\color{blue}\frac{3}{3}}\cdot\frac{n^2 + 4n-3}{(n+4)(n-3)} + \frac{4-3n}{3(n-4)}\cdot{\color{blue}\frac{n+4}{n+4}}$

. . $= \;\frac{3(n^2+4n-3) + (4-3n)(n+4)}{3(n+4)(n-4)} \;=\;\frac{3n^2 + 12n - 9 + 4n + 16 - 3n^2 -12n}{3(n+4)(n-4)}
$

. . $= \;\frac{4n+7}{3(n+4)(n-4)}$ . . . . There!