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

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Answer 1 · 2017-01-14T19:28:46

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

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

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.