Answer:
The values of a and b are [tex]$ \frac{-13}{7} $[/tex] and [tex]$ \frac{-2}{11} $[/tex] respectively.
Step-by-step explanation:
Given that [tex]$ \frac{x - 4}{x^2 + 7x - 18} = \frac{a}{x + 9} + \frac{b}{x - 2} $[/tex]
We solve this by partial fraction method.
Taking L.C.M. in the RHS, we get
[tex]$ \frac{x - 4}{x^2 + 7x - 18} = \frac{a(x - 2) + b(x + 9)}{(x + 9)(x - 2)} $[/tex]
[tex]$ \implies x - 4 = a(x - 2) + b(x + 9) $[/tex]
To find the value of 'b', substitute x = 2. This would make 'a' vanish leaving an equation with 'b'.
Therefore, 2 - 4 = a(2 - 2) + b(2 + 9)
⇒ -2 = 0 + b(11)
[tex]$ \implies b = \frac{-2}{11} $[/tex]
Similarly, substitute x = -9 to find the value of 'a'.
⇒ -9 - 4 = a(7) + b(0)
[tex]$ \implies a = \frac{-13}{7} $[/tex].
Therefore, the values of 'a' and 'b'b are: =-13/7 and -2/11 respectively.
