To find the partial fraction decomposition of the expression \(\frac{11x - 3}{(x - 2)(x + 3)}\), we can express it in the form:
\[ \frac{11x - 3}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} \]
where \(A\) and \(B\) are constants that we need to determine.
First, we can combine the right-hand side over a common denominator:
\[ \frac{A}{x - 2} + \frac{B}{x + 3} = \frac{A(x + 3) + B(x - 2)}{(x - 2)(x + 3)} \]
Setting the numerators equal gives us:
\[ 11x - 3 = A(x + 3) + B(x - 2) \]
Expanding the right side:
\[ A(x + 3) + B(x - 2) = Ax + 3A + Bx - 2B = (A + B)x + (3A - 2B) \]
Now we can set the coefficients for \(x\) and the constant terms equal:
- \(A + B = 11\) (coefficient of \(x\))
- \(3A - 2B = -3\) (constant term)
We now have a system of equations:
- \(A + B = 11\) (Equation 1)
- \(3A - 2B = -3\) (Equation 2)
We can solve this system step-by-step. From Equation 1, we can express \(B\) in terms of \(A\):
\[ B = 11 - A \]
Substituting this expression for \(B\) into Equation 2:
\[ 3A - 2(11 - A) = -3 \]
Simplifying this:
\[ 3A - 22 + 2A = -3 \] \[ 5A - 22 = -3 \] \[ 5A = 19 \] \[ A = \frac{19}{5} \]
Now, substituting \(A\) back into Equation 1 to find \(B\):
\[ \frac{19}{5} + B = 11 \] \[ B = 11 - \frac{19}{5} \] \[ B = \frac{55}{5} - \frac{19}{5} = \frac{36}{5} \]
Now we have both constants:
\[ A = \frac{19}{5}, \quad B = \frac{36}{5} \]
Finally, we can write the partial fraction decomposition:
\[ \frac{11x - 3}{(x - 2)(x + 3)} = \frac{\frac{19}{5}}{x - 2} + \frac{\frac{36}{5}}{x + 3} \]
Or, rewritten more cleanly:
\[ \frac{11x - 3}{(x - 2)(x + 3)} = \frac{19}{5(x - 2)} + \frac{36}{5(x + 3)} \]
This is the partial fraction decomposition of the given rational expression.
1 answer