An algebra problem by A Former Brilliant Member

To decompose the fraction, we factor the denominator and obtain,

$\frac{13}{(-2x + 3)(3x + 2)}$

Therefore we have $\frac{13}{(-2x + 3)(3x + 2)} =$ $\frac{A}{-2x + 3} +$ $\frac{B}{3x + 2}$

The equation above is an identity for all $x$ except $\frac{3}{2}$ and $\frac{-2}{3}$ . By multiplying both sides on the LCD we obtain,

$13 = A(3x + 2) + B(-2x + 3)$

We now find the constants $A$ and $B$ , substituting $\frac{3}{2}$ to the above equation, we get $A=2$

Substituting $\frac{-2}{3}$ , we get $B=3$

With these values for $A$ and $B$ ,

$\frac{13}{(-2x + 3)(3x + 2)} = -$ $\frac{2}{2x - 3} +$ $\frac{3}{3x + 2}$