Write as a single fraction and simplify,
(8/ x^2 + 2x -3) - (4/ x^2 - 1)
Can someone help me with this q please? Thanks
1. Factorize the denominators:
$\displaystyle \dfrac8{x^2+2x-3} - \dfrac4{x^2-1} = \dfrac8{(x-1)(x+3)} - \dfrac4{(x-1)(x+1)}$
2. The general denominator is (x-1)(x+1)(x+3).
3. Therefore the difference becomes:
$\displaystyle \dfrac{8(x+1)}{(x-1)(x+1)(x+3)} - \dfrac{4(x+3)}{(x-1)(x+1)(x+3)}$
4. Expand the numerators and collect like terms.
$\displaystyle \dfrac{8x+8 - 4x-12}{(x-1)(x+1)(x+3)} = \dfrac{4(x-1)}{(x-1)(x+1)(x+3)}$
Now cancel:
$\displaystyle \dfrac{4}{(x+1)(x+3)}\ ,\ x \neq 1$