Precalculus Examples

$\frac{4}{3 x (5 x + 1)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{5 x + 1}$

Step 1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $3 x (5 x + 1)$ .

$\frac{4 (3 x (5 x + 1))}{3 x (5 x + 1)} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.3

Cancel the common factor of $3$ .

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Step 1.3.1

Cancel the common factor.

$\frac{4 (3 x (5 x + 1))}{3 x (5 x + 1)} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.3.2

Rewrite the expression.

$\frac{4 (x (5 x + 1))}{x (5 x + 1)} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

$\frac{4 (x (5 x + 1))}{x (5 x + 1)} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.4.2

Rewrite the expression.

$\frac{4 (5 x + 1)}{5 x + 1} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.5

Cancel the common factor of $5 x + 1$ .

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Step 1.5.1

Cancel the common factor.

$\frac{4 (5 x + 1)}{5 x + 1} = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.5.2

Divide $4$ by $1$ .

$4 = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x$ .

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Step 1.6.1.1

Cancel the common factor.

$4 = \frac{A (3 x (5 x + 1))}{x} + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.1.2

Divide $A (3 (5 x + 1))$ by $1$ .

$4 = A (3 (5 x + 1)) + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.2

Rewrite using the commutative property of multiplication.

$4 = 3 A (5 x + 1) + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.3

Apply the distributive property.

$4 = 3 A (5 x) + 3 A \cdot 1 + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.4

Rewrite using the commutative property of multiplication.

$4 = 3 \cdot (5 A x) + 3 A \cdot 1 + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.5

Multiply $3$ by $1$ .

$4 = 3 \cdot (5 A x) + 3 A + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.6

Multiply $3$ by $5$ .

$4 = 15 A x + 3 A + \frac{(B) (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.7

Cancel the common factor of $5 x + 1$ .

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Step 1.6.7.1

Cancel the common factor.

$4 = 15 A x + 3 A + \frac{B (3 x (5 x + 1))}{5 x + 1}$

Step 1.6.7.2

Divide $(B) (3 x)$ by $1$ .

$4 = 15 A x + 3 A + (B) (3 x)$

Step 1.6.8

Rewrite using the commutative property of multiplication.

$4 = 15 A x + 3 A + 3 B x$

Step 1.7

Reorder.

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Step 1.7.1

Move $A$ .

$4 = 15 x A + 3 A + 3 B x$

Step 1.7.2

Move $B$ .

$4 = 15 x A + 3 A + 3 x B$

Step 1.7.3

Move $3 A$ .

$4 = 15 x A + 3 x B + 3 A$

Step 2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 15 A + 3 B$

Step 2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$4 = 3 A$

Step 2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = 15 A + 3 B$

$4 = 3 A$

$0 = 15 A + 3 B$

$4 = 3 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $4 = 3 A$ .

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Step 3.1.1

Rewrite the equation as $3 A = 4$ .

$3 A = 4$

$0 = 15 A + 3 B$

Step 3.1.2

Divide each term in $3 A = 4$ by $3$ and simplify.

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Step 3.1.2.1

Divide each term in $3 A = 4$ by $3$ .

$\frac{3 A}{3} = \frac{4}{3}$

$0 = 15 A + 3 B$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $3$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{3 A}{3} = \frac{4}{3}$

$0 = 15 A + 3 B$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{4}{3}$

$0 = 15 A + 3 B$

$A = \frac{4}{3}$

$0 = 15 A + 3 B$

$A = \frac{4}{3}$

$0 = 15 A + 3 B$

$A = \frac{4}{3}$

$0 = 15 A + 3 B$

$A = \frac{4}{3}$

$0 = 15 A + 3 B$

Step 3.2

Replace all occurrences of $A$ with $\frac{4}{3}$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $0 = 15 A + 3 B$ with $\frac{4}{3}$ .

$0 = 15 (\frac{4}{3}) + 3 B$

$A = \frac{4}{3}$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1.1

Factor $3$ out of $15$ .

$0 = 3 (5) (\frac{4}{3}) + 3 B$

$A = \frac{4}{3}$

Step 3.2.2.1.1.2

Cancel the common factor.

$0 = 3 \cdot (5 (\frac{4}{3})) + 3 B$

$A = \frac{4}{3}$

Step 3.2.2.1.1.3

Rewrite the expression.

$0 = 5 \cdot 4 + 3 B$

$A = \frac{4}{3}$

$0 = 5 \cdot 4 + 3 B$

$A = \frac{4}{3}$

Step 3.2.2.1.2

Multiply $5$ by $4$ .

$0 = 20 + 3 B$

$A = \frac{4}{3}$

$0 = 20 + 3 B$

$A = \frac{4}{3}$

$0 = 20 + 3 B$

$A = \frac{4}{3}$

$0 = 20 + 3 B$

$A = \frac{4}{3}$

Step 3.3

Solve for $B$ in $0 = 20 + 3 B$ .

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Step 3.3.1

Rewrite the equation as $20 + 3 B = 0$ .

$20 + 3 B = 0$

$A = \frac{4}{3}$

Step 3.3.2

Subtract $20$ from both sides of the equation.

$3 B = - 20$

$A = \frac{4}{3}$

Step 3.3.3

Divide each term in $3 B = - 20$ by $3$ and simplify.

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Step 3.3.3.1

Divide each term in $3 B = - 20$ by $3$ .

$\frac{3 B}{3} = \frac{- 20}{3}$

$A = \frac{4}{3}$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $3$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{3 B}{3} = \frac{- 20}{3}$

$A = \frac{4}{3}$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 20}{3}$

$A = \frac{4}{3}$

$B = \frac{- 20}{3}$

$A = \frac{4}{3}$

$B = \frac{- 20}{3}$

$A = \frac{4}{3}$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Move the negative in front of the fraction.

$B = - \frac{20}{3}$

$A = \frac{4}{3}$

$B = - \frac{20}{3}$

$A = \frac{4}{3}$

$B = - \frac{20}{3}$

$A = \frac{4}{3}$

$B = - \frac{20}{3}$

$A = \frac{4}{3}$

Step 3.4

Solve the system of equations.

$B = - \frac{20}{3} A = \frac{4}{3}$

Step 3.5

List all of the solutions.

$B = - \frac{20}{3}, A = \frac{4}{3}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{5 x + 1}$ with the values found for $A$ and $B$ .

$\frac{\frac{4}{3}}{x} + \frac{- \frac{20}{3}}{5 x + 1}$