# How do you decompose (-4y)/(3y^2-4y+1) into partial fractions?

Jan 14, 2017

$\frac{- 4 y}{3 {y}^{2} - 4 y + 1} = \frac{2}{3 y - 1} - \frac{2}{y - 1}$

#### Explanation:

Factor the denominator $\left(y - 1\right) \left(3 y - 1\right)$

Write the equation of the partial fractions with the unknowns, A and B:

$\frac{- 4 y}{\left(y - 1\right) \left(3 y - 1\right)} = \frac{A}{y - 1} + \frac{B}{3 y - 1}$

Multiply both sides by the denominator:

$- 4 y = A \left(3 y - 1\right) + B \left(y - 1\right)$

Make B disappear by setting y = 1:

$- 4 = A \left(3 - 1\right) + B \left(0\right)$

#A = -2

Substitute - 2 for A and set y = 0:

$- 4 \left(0\right) = - 2 \left(3 \left(0\right) - 1\right) + B \left(0 - 1\right)$

$B = 2$

Check:

$- \frac{2}{y - 1} + \frac{2}{3 y - 1}$

$- \frac{2}{y - 1} \frac{3 y - 1}{3 y - 1} + \frac{2}{3 y - 1} \frac{y - 1}{y - 1}$

$\frac{- 6 y + 2 + 2 y - 2}{\left(y - 1\right) \left(3 y - 1\right)}$

$\frac{- 4 y}{\left(y - 1\right) \left(3 y - 1\right)}$

This checks.