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Math  /  Algebra

QuestionFind the partial fraction decomposition. 20x−2(x−1)(3x−1)=\frac{20 x-2}{(x-1)(3 x-1)}= □\square

Studdy Solution

STEP 1

1. The given expression is a rational function that can be decomposed into simpler fractions.
2. The denominators of these simpler fractions are factors of the original denominator.
3. The original denominator is composed of linear factors that can be used to form the partial fractions.

STEP 2

1. Express the given fraction as a sum of partial fractions.
2. Set up equations by equating the numerators.
3. Solve the system of equations to find the constants in the partial fractions.
4. Substitute the constants back into the partial fractions.

STEP 3

Express the given fraction as a sum of partial fractions.
20x−2(x−1)(3x−1)=Ax−1+B3x−1 \frac{20x - 2}{(x-1)(3x-1)} = \frac{A}{x-1} + \frac{B}{3x-1}

STEP 4

Combine the right-hand side over a common denominator.
Ax−1+B3x−1=A(3x−1)+B(x−1)(x−1)(3x−1) \frac{A}{x-1} + \frac{B}{3x-1} = \frac{A(3x-1) + B(x-1)}{(x-1)(3x-1)}

STEP 5

Set the numerators equal to each other.
20x−2=A(3x−1)+B(x−1) 20x - 2 = A(3x-1) + B(x-1)

STEP 6

Expand and collect like terms on the right-hand side.
20x−2=3Ax−A+Bx−B 20x - 2 = 3Ax - A + Bx - B 20x−2=(3A+B)x−(A+B) 20x - 2 = (3A + B)x - (A + B)

STEP 7

Equate the coefficients of like terms from both sides of the equation.
For the coefficients of xx: 20=3A+B 20 = 3A + B
For the constant term: −2=−A−B -2 = -A - B

STEP 8

Solve the system of equations to find AA and BB.
Solve the first equation for BB: B=20−3A B = 20 - 3A
Substitute into the second equation: −2=−A−(20−3A) -2 = -A - (20 - 3A) −2=−A−20+3A -2 = -A - 20 + 3A −2=2A−20 -2 = 2A - 20 18=2A 18 = 2A A=9 A = 9

STEP 9

Substitute A=9A = 9 back into the expression for BB:
B=20−3(9) B = 20 - 3(9) B=20−27 B = 20 - 27 B=−7 B = -7

STEP 10

Substitute the values of AA and BB back into the partial fractions.
20x−2(x−1)(3x−1)=9x−1+−73x−1 \frac{20x - 2}{(x-1)(3x-1)} = \frac{9}{x-1} + \frac{-7}{3x-1}
Therefore, the partial fraction decomposition is:
20x−2(x−1)(3x−1)=9x−1−73x−1 \frac{20x - 2}{(x-1)(3x-1)} = \frac{9}{x-1} - \frac{7}{3x-1}

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