Precalculus Examples

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

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 $A$ .

$\frac{A}{x + 3}$

Step 1.2

$\frac{A}{x + 3} + \frac{B}{3 x - 1}$

Step 1.3

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

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

Step 1.4

Cancel the common factor of $x + 3$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Cancel the common factor of $3 x - 1$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Divide $4 x + 3$ by $1$ .

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

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x + 3$ .

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Step 1.6.1.1

Cancel the common factor.

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

Step 1.6.1.2

Divide $(A) (3 x - 1)$ by $1$ .

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

Step 1.6.2

Apply the distributive property.

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

Step 1.6.3

Rewrite using the commutative property of multiplication.

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

Step 1.6.4

Move $- 1$ to the left of $A$ .

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

Step 1.6.5

Rewrite $- 1 A$ as $- A$ .

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

Step 1.6.6

Cancel the common factor of $3 x - 1$ .

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Step 1.6.6.1

Cancel the common factor.

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

Step 1.6.6.2

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

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

Step 1.6.7

Apply the distributive property.

$4 x + 3 = 3 A x - A + B x + B \cdot 3$

Step 1.6.8

Move $3$ to the left of $B$ .

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

Step 1.7

Simplify the expression.

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Step 1.7.1

Move $A$ .

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

Step 1.7.2

Reorder $B$ and $x$ .

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

Step 1.7.3

Move $- A$ .

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

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.

$4 = 3 A + B$

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.

$3 = - 1 A + 3 B$

Step 2.3

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

$4 = 3 A + B$

$3 = - 1 A + 3 B$

$4 = 3 A + B$

$3 = - 1 A + 3 B$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

Rewrite the equation as $3 A + B = 4$ .

$3 A + B = 4$

$3 = - 1 A + 3 B$

Step 3.1.2

Subtract $3 A$ from both sides of the equation.

$B = 4 - 3 A$

$3 = - 1 A + 3 B$

$B = 4 - 3 A$

$3 = - 1 A + 3 B$

Step 3.2

Replace all occurrences of $B$ with $4 - 3 A$ in each equation.

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Step 3.2.1

Replace all occurrences of $B$ in $3 = - 1 A + 3 B$ with $4 - 3 A$ .

$3 = - 1 A + 3 (4 - 3 A)$

$B = 4 - 3 A$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 1 A + 3 (4 - 3 A)$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Rewrite $- 1 A$ as $- A$ .

$3 = - A + 3 (4 - 3 A)$

$B = 4 - 3 A$

Step 3.2.2.1.1.2

Apply the distributive property.

$3 = - A + 3 \cdot 4 + 3 (- 3 A)$

$B = 4 - 3 A$

Step 3.2.2.1.1.3

Multiply $3$ by $4$ .

$3 = - A + 12 + 3 (- 3 A)$

$B = 4 - 3 A$

Step 3.2.2.1.1.4

Multiply $- 3$ by $3$ .

$3 = - A + 12 - 9 A$

$B = 4 - 3 A$

$3 = - A + 12 - 9 A$

$B = 4 - 3 A$

Step 3.2.2.1.2

Subtract $9 A$ from $- A$ .

$3 = - 10 A + 12$

$B = 4 - 3 A$

$3 = - 10 A + 12$

$B = 4 - 3 A$

$3 = - 10 A + 12$

$B = 4 - 3 A$

$3 = - 10 A + 12$

$B = 4 - 3 A$

Step 3.3

Solve for $A$ in $3 = - 10 A + 12$ .

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Step 3.3.1

Rewrite the equation as $- 10 A + 12 = 3$ .

$- 10 A + 12 = 3$

$B = 4 - 3 A$

Step 3.3.2

Move all terms not containing $A$ to the right side of the equation.

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Step 3.3.2.1

Subtract $12$ from both sides of the equation.

$- 10 A = 3 - 12$

$B = 4 - 3 A$

Step 3.3.2.2

Subtract $12$ from $3$ .

$- 10 A = - 9$

$B = 4 - 3 A$

$- 10 A = - 9$

$B = 4 - 3 A$

Step 3.3.3

Divide each term in $- 10 A = - 9$ by $- 10$ and simplify.

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Step 3.3.3.1

Divide each term in $- 10 A = - 9$ by $- 10$ .

$\frac{- 10 A}{- 10} = \frac{- 9}{- 10}$

$B = 4 - 3 A$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 10$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 10 A}{- 10} = \frac{- 9}{- 10}$

$B = 4 - 3 A$

Step 3.3.3.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 9}{- 10}$

$B = 4 - 3 A$

$A = \frac{- 9}{- 10}$

$B = 4 - 3 A$

$A = \frac{- 9}{- 10}$

$B = 4 - 3 A$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Dividing two negative values results in a positive value.

$A = \frac{9}{10}$

$B = 4 - 3 A$

$A = \frac{9}{10}$

$B = 4 - 3 A$

$A = \frac{9}{10}$

$B = 4 - 3 A$

$A = \frac{9}{10}$

$B = 4 - 3 A$

Step 3.4

Replace all occurrences of $A$ with $\frac{9}{10}$ in each equation.

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Step 3.4.1

Replace all occurrences of $A$ in $B = 4 - 3 A$ with $\frac{9}{10}$ .

$B = 4 - 3 (\frac{9}{10})$

$A = \frac{9}{10}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $4 - 3 (\frac{9}{10})$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Multiply $- 3 (\frac{9}{10})$ .

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Step 3.4.2.1.1.1.1

Combine $- 3$ and $\frac{9}{10}$ .

$B = 4 + \frac{- 3 \cdot 9}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.1.1.2

Multiply $- 3$ by $9$ .

$B = 4 + \frac{- 27}{10}$

$A = \frac{9}{10}$

$B = 4 + \frac{- 27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.1.2

Move the negative in front of the fraction.

$B = 4 - \frac{27}{10}$

$A = \frac{9}{10}$

$B = 4 - \frac{27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$B = 4 \cdot \frac{10}{10} - \frac{27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.3

Combine $4$ and $\frac{10}{10}$ .

$B = \frac{4 \cdot 10}{10} - \frac{27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.4

Combine the numerators over the common denominator.

$B = \frac{4 \cdot 10 - 27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.5

Simplify the numerator.

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Step 3.4.2.1.5.1

Multiply $4$ by $10$ .

$B = \frac{40 - 27}{10}$

$A = \frac{9}{10}$

Step 3.4.2.1.5.2

Subtract $27$ from $40$ .

$B = \frac{13}{10}$

$A = \frac{9}{10}$

$B = \frac{13}{10}$

$A = \frac{9}{10}$

$B = \frac{13}{10}$

$A = \frac{9}{10}$

$B = \frac{13}{10}$

$A = \frac{9}{10}$

$B = \frac{13}{10}$

$A = \frac{9}{10}$

Step 3.5

List all of the solutions.

$B = \frac{13}{10}, A = \frac{9}{10}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x + 3} + \frac{B}{3 x - 1}$ with the values found for $A$ and $B$ .

$\frac{\frac{9}{10}}{x + 3} + \frac{\frac{13}{10}}{3 x - 1}$