Question

6x-1/((x-3)(x+1)) into partial fractions

Answer 1

$\frac{6x-1}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}=\frac{A(x+1)+B(x-3)}{(x-3)(x+1)}=\frac{Ax+A+Bx-3B}{(x-3)(x+1)}=\frac{x(A+B)+A-3B}{(x-3)(x+1)}$

$\begin{cases} A+B=6 \\ A-3B=-1 \end{cases}$

$\begin{cases} A+B=6 \\ A=3B-1 \end{cases}$

$3B-1+B=6$

$4B=7$

$B=\frac{7}{4}$

$A=3\cdot\frac{7}{4}-1=\frac{17}{4}$

$\frac{6x-1}{(x-3)(x+1)}=\frac{17}{4(x-3)}+\frac{7}{4(x+1)}$