Precalculus Examples

$\frac{x}{(3 x - 7) (2 x + 1)}$

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 $A$ .

$\frac{A}{3 x - 7}$

Step 1.2

$\frac{A}{3 x - 7} + \frac{B}{2 x + 1}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(3 x - 7) (2 x + 1)$ .

$\frac{x (3 x - 7) (2 x + 1)}{(3 x - 7) (2 x + 1)} = \frac{(A) (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.4

Cancel the common factor of $3 x - 7$ .

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Step 1.4.1

Cancel the common factor.

$\frac{x (3 x - 7) (2 x + 1)}{(3 x - 7) (2 x + 1)} = \frac{(A) (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.4.2

Rewrite the expression.

$\frac{x (2 x + 1)}{2 x + 1} = \frac{(A) (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.5

Cancel the common factor of $2 x + 1$ .

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Step 1.5.1

Cancel the common factor.

$\frac{x (2 x + 1)}{2 x + 1} = \frac{(A) (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.5.2

Divide $x$ by $1$ .

$x = \frac{(A) (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $3 x - 7$ .

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Step 1.6.1.1

Cancel the common factor.

$x = \frac{A (3 x - 7) (2 x + 1)}{3 x - 7} + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.1.2

Divide $(A) (2 x + 1)$ by $1$ .

$x = (A) (2 x + 1) + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.2

Apply the distributive property.

$x = A (2 x) + A \cdot 1 + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.3

Rewrite using the commutative property of multiplication.

$x = 2 A x + A \cdot 1 + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.4

Multiply $A$ by $1$ .

$x = 2 A x + A + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.5

Cancel the common factor of $2 x + 1$ .

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Step 1.6.5.1

Cancel the common factor.

$x = 2 A x + A + \frac{(B) (3 x - 7) (2 x + 1)}{2 x + 1}$

Step 1.6.5.2

Divide $(B) (3 x - 7)$ by $1$ .

$x = 2 A x + A + (B) (3 x - 7)$

Step 1.6.6

Apply the distributive property.

$x = 2 A x + A + B (3 x) + B \cdot - 7$

Step 1.6.7

Rewrite using the commutative property of multiplication.

$x = 2 A x + A + 3 B x + B \cdot - 7$

Step 1.6.8

Move $- 7$ to the left of $B$ .

$x = 2 A x + A + 3 B x - 7 B$

Step 1.7

Reorder.

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Step 1.7.1

Move $A$ .

$x = 2 x A + A + 3 B x - 7 B$

Step 1.7.2

Move $B$ .

$x = 2 x A + A + 3 x B - 7 B$

Step 1.7.3

Move $A$ .

$x = 2 x A + 3 x B + A - 7 B$

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.

$1 = 2 A + 3 B$

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.

$0 = A - 7 B$

Step 2.3

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

$1 = 2 A + 3 B$

$0 = A - 7 B$

$1 = 2 A + 3 B$

$0 = A - 7 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $0 = A - 7 B$ .

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Step 3.1.1

Rewrite the equation as $A - 7 B = 0$ .

$A - 7 B = 0$

$1 = 2 A + 3 B$

Step 3.1.2

Add $7 B$ to both sides of the equation.

$A = 7 B$

$1 = 2 A + 3 B$

$A = 7 B$

$1 = 2 A + 3 B$

Step 3.2

Replace all occurrences of $A$ with $7 B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $1 = 2 A + 3 B$ with $7 B$ .

$1 = 2 (7 B) + 3 B$

$A = 7 B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (7 B) + 3 B$ .

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Step 3.2.2.1.1

Multiply $7$ by $2$ .

$1 = 14 B + 3 B$

$A = 7 B$

Step 3.2.2.1.2

Add $14 B$ and $3 B$ .

$1 = 17 B$

$A = 7 B$

$1 = 17 B$

$A = 7 B$

$1 = 17 B$

$A = 7 B$

$1 = 17 B$

$A = 7 B$

Step 3.3

Solve for $B$ in $1 = 17 B$ .

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Step 3.3.1

Rewrite the equation as $17 B = 1$ .

$17 B = 1$

$A = 7 B$

Step 3.3.2

Divide each term in $17 B = 1$ by $17$ and simplify.

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Step 3.3.2.1

Divide each term in $17 B = 1$ by $17$ .

$\frac{17 B}{17} = \frac{1}{17}$

$A = 7 B$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $17$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{17 B}{17} = \frac{1}{17}$

$A = 7 B$

Step 3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{17}$

$A = 7 B$

$B = \frac{1}{17}$

$A = 7 B$

$B = \frac{1}{17}$

$A = 7 B$

$B = \frac{1}{17}$

$A = 7 B$

$B = \frac{1}{17}$

$A = 7 B$

Step 3.4

Replace all occurrences of $B$ with $\frac{1}{17}$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = 7 B$ with $\frac{1}{17}$ .

$A = 7 (\frac{1}{17})$

$B = \frac{1}{17}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Combine $7$ and $\frac{1}{17}$ .

$A = \frac{7}{17}$

$B = \frac{1}{17}$

$A = \frac{7}{17}$

$B = \frac{1}{17}$

$A = \frac{7}{17}$

$B = \frac{1}{17}$

Step 3.5

List all of the solutions.

$A = \frac{7}{17}, B = \frac{1}{17}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{3 x - 7} + \frac{B}{2 x + 1}$ with the values found for $A$ and $B$ .

$\frac{\frac{7}{17}}{3 x - 7} + \frac{\frac{1}{17}}{2 x + 1}$