Please walk me through how to solve these:
1. [w/(w-1)] + w = [(4w-3)/(w-1)]
2. {(4n^2)/[(n^2)-9]} - [2n/(n+3)] = 3/(n-3)
3. [4/(z-2)] - [(z+6)/(z+1)] = 1
Thanks in advance =)
1.
$\displaystyle \frac{w}{(w-1)} + w = \frac{(4w-3)}{(w-1)}$
$\displaystyle \frac{w(w-1)}{(w-1)} + w = \frac{(4w-3)(w-1)}{(w-1)}$
$\displaystyle w + w = (4w-3)$
$\displaystyle 2w = 4w-3$
$\displaystyle 3 = 2w$
$\displaystyle w = \frac{2}{3}$
2.
$\displaystyle \frac{4n^2}{n^2-9} - \frac{2n}{(n+3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{4n^2}{(n+3)(n-3)} - \frac{2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{4n^2-2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{4n^2-2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{2n(n+3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{2n}{(n-3)} = \frac{3}{(n-3)}$
$\displaystyle \frac{2n(n-3)}{(n-3)} = \frac{3(n-3)}{(n-3)}$
$\displaystyle 2n = 3$
$\displaystyle n = \frac{2}{3}$