Question

o). express $\frac{7x}{x^{2}-7x + 10}$ as a sum of its partial fractions.

Explanation:

Step1: Factor the denominator

First, factor $x^{2}-7x + 10=(x - 2)(x - 5)$. Then we assume $\frac{7x}{x^{2}-7x + 10}=\frac{A}{x - 2}+\frac{B}{x - 5}$.

Step2: Combine the right - hand side

$\frac{A}{x - 2}+\frac{B}{x - 5}=\frac{A(x - 5)+B(x - 2)}{(x - 2)(x - 5)}=\frac{(A + B)x+(-5A - 2B)}{(x - 2)(x - 5)}$.

Step3: Equate the numerators

We have the system of equations

$$\begin{cases}A + B=7\\-5A-2B = 0\end{cases}$$

. From the first equation $A=7 - B$. Substitute it into the second equation: $-5(7 - B)-2B=0$. Expand to get $-35 + 5B-2B=0$, then $3B=35$, so $B=\frac{35}{3}$. And $A=7-\frac{35}{3}=\frac{21 - 35}{3}=-\frac{14}{3}$.

Answer:

$-\frac{14}{3(x - 2)}+\frac{35}{3(x - 5)}$