Answer
$ \displaystyle \frac{24}{7(x-4)}+\frac{25}{7(x+3)}$
Work Step by Step
First, factor the trinomial in the denominator
(Two factors of -12 that add to -1 are -4 and 3)
$\displaystyle \frac{7x-4}{(x-4)(x+3)}=...$
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{7x-4}{(x-4)(x+3)}=\frac{A}{x-4}+\frac{B}{x+3}$
multiply both sides with $(x-4)(x+3)\quad $(LCD)
$7x-4=A(x+3)+B(x-4)$
$ 7x-4=Ax+3A+Bx-4B\qquad$ /group like terms...
$7x-4=x(A+B)+(3A-4B)$
... we now equate the coefficients of like powers...
$\left\{\begin{array}{ll}
I. & 7=A+B\\
II. & -4=3A-4B
\end{array}\right.\qquad $solve the system
$(II-3\times I)$ eliminates $A...$
$-4-21=3A-3A-4B-3B$
$-25=-7B$
$B=\displaystyle \frac{25}{7}$
back substitute: $7=A+B$
$7=A+\displaystyle \frac{25}{7}$
$A=\displaystyle \frac{49-25}{7}=\frac{24}{7}$
Rewrite the setup decomposition with $A=\displaystyle \frac{24}{7}, B=\frac{25}{7}$
$\displaystyle \frac{7x-4}{(x-4)(x+3)}=\frac{\frac{24}{7}}{x-4}+\frac{\frac{25}{7}}{x+3}=$
$= \displaystyle \frac{24}{7(x-4)}+\frac{25}{7(x+3)}$
