Precalculus Examples

$\frac{15 x^{2} - 11 x - 5}{x (x + 1) (2 x - 5)}$

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 $B$ .

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

Step 1.2

$\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{2 x - 5}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x + 1) (2 x - 5)$ .

$\frac{(15 x^{2} - 11 x - 5) (x (x + 1) (2 x - 5))}{x (x + 1) (2 x - 5)} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(15 x^{2} - 11 x - 5) (x (x + 1) (2 x - 5))}{x (x + 1) (2 x - 5)} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.4.2

Rewrite the expression.

$\frac{(15 x^{2} - 11 x - 5) ((x + 1) (2 x - 5))}{(x + 1) (2 x - 5)} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.5

Cancel the common factor of $x + 1$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(15 x^{2} - 11 x - 5) ((x + 1) (2 x - 5))}{(x + 1) (2 x - 5)} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.5.2

Rewrite the expression.

$\frac{(15 x^{2} - 11 x - 5) (2 x - 5)}{2 x - 5} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.6

Cancel the common factor of $2 x - 5$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(15 x^{2} - 11 x - 5) (2 x - 5)}{2 x - 5} = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.6.2

Divide $15 x^{2} - 11 x - 5$ by $1$ .

$15 x^{2} - 11 x - 5 = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x$ .

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Step 1.7.1.1

Cancel the common factor.

$15 x^{2} - 11 x - 5 = \frac{A (x (x + 1) (2 x - 5))}{x} + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.1.2

Divide $A ((x + 1) (2 x - 5))$ by $1$ .

$15 x^{2} - 11 x - 5 = A ((x + 1) (2 x - 5)) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.2

Expand $(x + 1) (2 x - 5)$ using the FOIL Method.

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Step 1.7.2.1

Apply the distributive property.

$15 x^{2} - 11 x - 5 = A (x (2 x - 5) + 1 (2 x - 5)) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.2.2

Apply the distributive property.

$15 x^{2} - 11 x - 5 = A (x (2 x) + x \cdot - 5 + 1 (2 x - 5)) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.2.3

Apply the distributive property.

$15 x^{2} - 11 x - 5 = A (x (2 x) + x \cdot - 5 + 1 (2 x) + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3

Simplify and combine like terms.

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Step 1.7.3.1

Simplify each term.

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Step 1.7.3.1.1

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = A (2 x \cdot x + x \cdot - 5 + 1 (2 x) + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.7.3.1.2.1

Move $x$ .

$15 x^{2} - 11 x - 5 = A (2 (x \cdot x) + x \cdot - 5 + 1 (2 x) + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.1.2.2

Multiply $x$ by $x$ .

$15 x^{2} - 11 x - 5 = A (2 x^{2} + x \cdot - 5 + 1 (2 x) + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.1.3

Move $- 5$ to the left of $x$ .

$15 x^{2} - 11 x - 5 = A (2 x^{2} - 5 \cdot x + 1 (2 x) + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.1.4

Multiply $2 x$ by $1$ .

$15 x^{2} - 11 x - 5 = A (2 x^{2} - 5 x + 2 x + 1 \cdot - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.1.5

Multiply $- 5$ by $1$ .

$15 x^{2} - 11 x - 5 = A (2 x^{2} - 5 x + 2 x - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.3.2

Add $- 5 x$ and $2 x$ .

$15 x^{2} - 11 x - 5 = A (2 x^{2} - 3 x - 5) + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.4

Apply the distributive property.

$15 x^{2} - 11 x - 5 = A (2 x^{2}) + A (- 3 x) + A \cdot - 5 + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.5

Simplify.

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Step 1.7.5.1

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = 2 A x^{2} + A (- 3 x) + A \cdot - 5 + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.5.2

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x + A \cdot - 5 + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.5.3

Move $- 5$ to the left of $A$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 \cdot A + \frac{(B) (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.6

Cancel the common factor of $x + 1$ .

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Step 1.7.6.1

Cancel the common factor.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + \frac{B (x (x + 1) (2 x - 5))}{x + 1} + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.6.2

Divide $(B) (x (2 x - 5))$ by $1$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + (B) (x (2 x - 5)) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.7

Apply the distributive property.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (x (2 x) + x \cdot - 5) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.8

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 x \cdot x + x \cdot - 5) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.9

Move $- 5$ to the left of $x$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 x \cdot x - 5 \cdot x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.10

Multiply $x$ by $x$ by adding the exponents.

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Step 1.7.10.1

Move $x$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 (x \cdot x) - 5 \cdot x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.10.2

Multiply $x$ by $x$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 x^{2} - 5 \cdot x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 x^{2} - 5 x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.11

Apply the distributive property.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + B (2 x^{2}) + B (- 5 x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.12

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} + B (- 5 x) + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.13

Rewrite using the commutative property of multiplication.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + \frac{(C) (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.14

Cancel the common factor of $2 x - 5$ .

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Step 1.7.14.1

Cancel the common factor.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + \frac{C (x (x + 1) (2 x - 5))}{2 x - 5}$

Step 1.7.14.2

Divide $(C) (x (x + 1))$ by $1$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + (C) (x (x + 1))$

Step 1.7.15

Apply the distributive property.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + C (x \cdot x + x \cdot 1)$

Step 1.7.16

Multiply $x$ by $x$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + C (x^{2} + x \cdot 1)$

Step 1.7.17

Multiply $x$ by $1$ .

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + C (x^{2} + x)$

Step 1.7.18

Apply the distributive property.

$15 x^{2} - 11 x - 5 = 2 A x^{2} - 3 A x - 5 A + 2 B x^{2} - 5 B x + C x^{2} + C x$

Step 1.8

Reorder.

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Step 1.8.1

Move $A$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 A x - 5 A + 2 B x^{2} - 5 B x + C x^{2} + C x$

Step 1.8.2

Move $A$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 x A - 5 A + 2 B x^{2} - 5 B x + C x^{2} + C x$

Step 1.8.3

Move $B$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 x A - 5 A + 2 x^{2} B - 5 B x + C x^{2} + C x$

Step 1.8.4

Move $B$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 x A - 5 A + 2 x^{2} B - 5 x B + C x^{2} + C x$

Step 1.8.5

Move $- 5 A$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 x A + 2 x^{2} B - 5 x B + C x^{2} + C x - 5 A$

Step 1.8.6

Move $- 5 x B$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A - 3 x A + 2 x^{2} B + C x^{2} - 5 x B + C x - 5 A$

Step 1.8.7

Move $- 3 x A$ .

$15 x^{2} - 11 x - 5 = 2 x^{2} A + 2 x^{2} B + C x^{2} - 3 x A - 5 x B + C x - 5 A$

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$15 = 2 A + 2 B + C$

Step 2.2

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.

$- 11 = - 3 A - 5 B + C$

Step 2.3

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.

$- 5 = - 5 A$

Step 2.4

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

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$- 5 = - 5 A$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$- 5 = - 5 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $- 5 = - 5 A$ .

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Step 3.1.1

Rewrite the equation as $- 5 A = - 5$ .

$- 5 A = - 5$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

Step 3.1.2

Divide each term in $- 5 A = - 5$ by $- 5$ and simplify.

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Step 3.1.2.1

Divide each term in $- 5 A = - 5$ by $- 5$ .

$\frac{- 5 A}{- 5} = \frac{- 5}{- 5}$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $- 5$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{- 5 A}{- 5} = \frac{- 5}{- 5}$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 5}{- 5}$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$A = \frac{- 5}{- 5}$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$A = \frac{- 5}{- 5}$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $- 5$ by $- 5$ .

$A = 1$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$A = 1$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$A = 1$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

$A = 1$

$15 = 2 A + 2 B + C$

$- 11 = - 3 A - 5 B + C$

Step 3.2

Replace all occurrences of $A$ with $1$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $15 = 2 A + 2 B + C$ with $1$ .

$15 = 2 (1) + 2 B + C$

$A = 1$

$- 11 = - 3 A - 5 B + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Multiply $2$ by $1$ .

$15 = 2 + 2 B + C$

$A = 1$

$- 11 = - 3 A - 5 B + C$

$15 = 2 + 2 B + C$

$A = 1$

$- 11 = - 3 A - 5 B + C$

Step 3.2.3

Replace all occurrences of $A$ in $- 11 = - 3 A - 5 B + C$ with $1$ .

$- 11 = - 3 \cdot 1 - 5 B + C$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Multiply $- 3$ by $1$ .

$- 11 = - 3 - 5 B + C$

$15 = 2 + 2 B + C$

$A = 1$

$- 11 = - 3 - 5 B + C$

$15 = 2 + 2 B + C$

$A = 1$

$- 11 = - 3 - 5 B + C$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.3

Solve for $C$ in $- 11 = - 3 - 5 B + C$ .

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Step 3.3.1

Rewrite the equation as $- 3 - 5 B + C = - 11$ .

$- 3 - 5 B + C = - 11$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.3.2

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

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Step 3.3.2.1

Add $3$ to both sides of the equation.

$- 5 B + C = - 11 + 3$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.3.2.2

Add $5 B$ to both sides of the equation.

$C = - 11 + 3 + 5 B$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.3.2.3

Add $- 11$ and $3$ .

$C = - 8 + 5 B$

$15 = 2 + 2 B + C$

$A = 1$

$C = - 8 + 5 B$

$15 = 2 + 2 B + C$

$A = 1$

$C = - 8 + 5 B$

$15 = 2 + 2 B + C$

$A = 1$

Step 3.4

Replace all occurrences of $C$ with $- 8 + 5 B$ in each equation.

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Step 3.4.1

Replace all occurrences of $C$ in $15 = 2 + 2 B + C$ with $- 8 + 5 B$ .

$15 = 2 + 2 B - 8 + 5 B$

$C = - 8 + 5 B$

$A = 1$

Step 3.4.2

Simplify $15 = 2 + 2 B - 8 + 5 B$ .

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Step 3.4.2.1

Simplify the left side.

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Step 3.4.2.1.1

Remove parentheses.

$15 = 2 + 2 B - 8 + 5 B$

$C = - 8 + 5 B$

$A = 1$

$15 = 2 + 2 B - 8 + 5 B$

$C = - 8 + 5 B$

$A = 1$

Step 3.4.2.2

Simplify the right side.

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Step 3.4.2.2.1

Simplify $2 + 2 B - 8 + 5 B$ .

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Step 3.4.2.2.1.1

Subtract $8$ from $2$ .

$15 = 2 B - 6 + 5 B$

$C = - 8 + 5 B$

$A = 1$

Step 3.4.2.2.1.2

Add $2 B$ and $5 B$ .

$15 = 7 B - 6$

$C = - 8 + 5 B$

$A = 1$

$15 = 7 B - 6$

$C = - 8 + 5 B$

$A = 1$

$15 = 7 B - 6$

$C = - 8 + 5 B$

$A = 1$

$15 = 7 B - 6$

$C = - 8 + 5 B$

$A = 1$

$15 = 7 B - 6$

$C = - 8 + 5 B$

$A = 1$

Step 3.5

Solve for $B$ in $15 = 7 B - 6$ .

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Step 3.5.1

Rewrite the equation as $7 B - 6 = 15$ .

$7 B - 6 = 15$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.2

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

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Step 3.5.2.1

Add $6$ to both sides of the equation.

$7 B = 15 + 6$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.2.2

Add $15$ and $6$ .

$7 B = 21$

$C = - 8 + 5 B$

$A = 1$

$7 B = 21$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.3

Divide each term in $7 B = 21$ by $7$ and simplify.

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Step 3.5.3.1

Divide each term in $7 B = 21$ by $7$ .

$\frac{7 B}{7} = \frac{21}{7}$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $7$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{7 B}{7} = \frac{21}{7}$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{21}{7}$

$C = - 8 + 5 B$

$A = 1$

$B = \frac{21}{7}$

$C = - 8 + 5 B$

$A = 1$

$B = \frac{21}{7}$

$C = - 8 + 5 B$

$A = 1$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $21$ by $7$ .

$B = 3$

$C = - 8 + 5 B$

$A = 1$

$B = 3$

$C = - 8 + 5 B$

$A = 1$

$B = 3$

$C = - 8 + 5 B$

$A = 1$

$B = 3$

$C = - 8 + 5 B$

$A = 1$

Step 3.6

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

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Step 3.6.1

Replace all occurrences of $B$ in $C = - 8 + 5 B$ with $3$ .

$C = - 8 + 5 (3)$

$B = 3$

$A = 1$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- 8 + 5 (3)$ .

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Step 3.6.2.1.1

Multiply $5$ by $3$ .

$C = - 8 + 15$

$B = 3$

$A = 1$

Step 3.6.2.1.2

Add $- 8$ and $15$ .

$C = 7$

$B = 3$

$A = 1$

$C = 7$

$B = 3$

$A = 1$

$C = 7$

$B = 3$

$A = 1$

$C = 7$

$B = 3$

$A = 1$

Step 3.7

List all of the solutions.

$C = 7, B = 3, A = 1$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{2 x - 5}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{1}{x} + \frac{3}{x + 1} + \frac{7}{2 x - 5}$