Answer:
[tex]\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}[/tex]
Step-by-step explanation:
The given rational expression is:
[tex]\frac{3x+4}{x^{2}-6x+5} = \frac{3x+4}{(x-1)(x-5)}[/tex]
We can use concept of Partial Fractions to solve this problem. Let,
[tex]\frac{3x+4}{(x-1)(x-5)}=\frac{A}{x-1} +\frac{B}{x-5}[/tex]
Multiplying both sides by (x - 1)(x - 5), we get:
[tex]3x+4=A(x-5)+B(x-1)[/tex]
Substituting x = 5, we get:
[tex]3(5)+4=A(5-5)+B(5-1)\\\\ 15+4=0+4B\\\\ 19=4B\\\\ B=\frac{19}{4}[/tex]
Substituting x = 1, we get:
[tex]3(1)+4=A(1-5)+B(1-1)\\\\ 7=-4A\\\\ A=-\frac{7}{4}[/tex]
Substituting the value of A and B, back in the original equation, we get:
[tex]\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}[/tex]