Question

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

Answer 1

Please see the explanation for the process. The answer is:

`(-4y)/(3y^2 - 4y + 1) = 2/(3y - 1)-2/(y - 1)`

Explanation:

Factor the denominator `(y - 1)(3y - 1)`

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

`(-4y)/((y - 1)(3y - 1)) = A/(y - 1) + B/(3y - 1)`

Multiply both sides by the denominator:

`-4y = A(3y - 1) + B(y - 1)`

Make B disappear by setting y = 1:

`-4 = A(3 - 1) + B(0)`

#A = -2

Substitute - 2 for A and set y = 0:

`-4(0) = -2(3(0) - 1) + B(0 - 1)`

`B = 2`

Check:

`-2/(y - 1) + 2/(3y - 1)`

`-2/(y - 1)(3y - 1)/(3y - 1) + 2/(3y - 1)(y - 1)/(y - 1)`

`(-6y + 2 + 2y - 2)/((y - 1)(3y - 1))`

`(-4y)/((y - 1)(3y - 1))`

This checks.