# Question #c311f

May 5, 2016

$\frac{A}{B} = \frac{5}{3}$

We can find exact values of $A$ and $B$:
$A = \frac{3}{19}$, $B = \frac{5}{19}$

#### Explanation:

Given: $\frac{1}{\left(3 - 5 x\right) \left(2 + 3 x\right)} = \frac{A}{3 - 5 x} + \frac{B}{2 + 3 x}$

Bring the right side to common denominator:
$\frac{A}{3 - 5 x} + \frac{B}{2 + 3 x} = \frac{A \left(2 + 3 x\right) + B \left(3 - 5 x\right)}{\left(3 - 5 x\right) \left(2 + 3 x\right)}$

Compare this with a given equality.
Obviously,
$1 = A \left(2 + 3 x\right) + B \left(3 - 5 x\right)$ for any $x$.
Equivalently,
$\left(3 A - 5 B\right) x + 2 A + 3 B - 1 = 0$

Since this equality should hold for all $x$, the coefficient at $x$ must be equal to zero, that is, $3 A - 5 B = 0$ and the free member should also be equal to zero, that is, $2 A + 3 B - 1 = 0$.

From the first equality, we conclude that $\frac{A}{B} = \frac{5}{3}$.
Using this ratio and the second equality, we can find $A$ and $B$:
$2 \cdot \frac{5}{3} B + 3 B - 1 = 0$
$10 B + 9 B - 3 = 0$
$B = \frac{3}{19}$
$A = \frac{5}{19}$

CHECK:
$\frac{5}{19} \left(2 + 3 x\right) + \frac{3}{19} \left(3 - 5 x\right) = \frac{10 + 15 x + 9 - 15 x}{19} = 1$
as it should.