Answer:
1) The values of 'a' and 'b' are [tex]$ \frac{3}{4} $[/tex] and [tex]$ \frac{3}{4} $[/tex] respectively.
2) The values of 'a' and 'b' are '3' and '-3' respectively.
Step-by-step explanation:
1) Given: [tex]$ \frac{3x}{(x - 2)(3x + 2)} = \frac{a}{x - 2} + \frac{b}{3x + 2} $[/tex]
We solve this by partial fraction method.
Taking LCM in the RHS we get,
[tex]$ \frac{3x}{(x - 2)(3x + 2)} = \frac{a(3x + 2) + b(x - 2)}{(x - 2)(3x + 2)} $[/tex]
[tex]$ \implies 3x = a(3x + 2) + b(x - 2) $[/tex]
To find the value of 'a', substitute x = 2. This would make 'b' vanish leaving an equation with 'a'.
[tex]$ \therefore 3(2) = a(3.2 + 2) + b (2 - 2) \implies 6 = a(8) $[/tex]
[tex]$ \implies a = \frac{-2}{3} $[/tex]
Now, Substitute [tex]$ x = \frac{-2}{3} $[/tex] to solve for 'b'.
[tex]$ \implies 3(\frac{-2}{3}) = a (3.\frac{-2}{3} + 2) + b(\frac{-2}{3} -2) $[/tex]
[tex]$ \implies -2 = b \frac{-8}{3} $[/tex]
[tex]$ \implies b = \frac{3}{4} $[/tex]
Therefore, a = [tex]$ \frac{3}{4} $[/tex] and b = [tex]$ \frac{3}{4} $[/tex]
2) Given [tex]$ \frac{3}{x^2 - 5x + 6} = \frac{a}{x - 2} + \frac{b}{x - 3} $[/tex]
We follow the same procedure as (1).
Taking LCM we get
[tex]$ \frac{3}{x^2 - 5x + 6} = \frac{a (x - 3) + b(x - 2)}{(x^2 - 5x + 6)} $[/tex]
[tex]$ \implies 3 = a(x - 3) + b(x - 2) $[/tex]
Substituting x = 2, we get:
3 = a(-1) [tex]$ \implies a = -3 $[/tex]
Also, Substituting x = 3, we get:
3 = b(1)
[tex]$ \implies b = 1 $[/tex]
Therefore, the values of a and b are -1 and 1 respectively.
