1. ## Rational Equations

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

2. Originally Posted by TWN
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

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.

$\displaystyle \frac{4n^2}{n^2-9} - \frac{2n}{(n+3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{4n^2}{(n+3)(n-3)} - \frac{2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{4n^2-2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{4n^2-2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{2n(n+3)}{(n+3)(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{2n}{(n-3)} = \frac{3}{(n-3)}$

$\displaystyle \frac{2n(n-3)}{(n-3)} = \frac{3(n-3)}{(n-3)}$

$\displaystyle 2n = 3$

$\displaystyle n = \frac{2}{3}$

3. Thank you for the quick reply!
First of all, if 2x=3, the solution would be 3/2, not 2/3. This applies to both solutions you provided.
I tried plugging both 2/3 and 3/2 into both original equations and neither worked. Something seems to be incorrect.