## College Algebra (6th Edition)

$\displaystyle \frac{24}{7(x-4)}+\frac{25}{7(x+3)}$
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 $A=\displaystyle \frac{24}{7}, B=\frac{25}{7}$ $\displaystyle \frac{7x-4}{(x-4)(x+3)}=\frac{\frac{24}{7}}{x-4}+\frac{\frac{25}{7}}{x+3}=$ $= \displaystyle \frac{24}{7(x-4)}+\frac{25}{7(x+3)}$