Algebra Examples

$\frac{4 x^{2} + 55 x + 25}{5 x {(3 x + 5)}^{2}}$

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 $B$ .

$\frac{A}{x} + \frac{B}{{(3 x + 5)}^{2}}$

Step 1.2

$\frac{A}{x} + \frac{B}{{(3 x + 5)}^{2}} + \frac{C}{3 x + 5}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $5 x {(3 x + 5)}^{2}$ .

$\frac{(4 x^{2} + 55 x + 25) (5 x {(3 x + 5)}^{2})}{5 x {(3 x + 5)}^{2}} = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.4

Cancel the common factor of $5$ .

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Step 1.4.1

Cancel the common factor.

Step 1.4.2

Rewrite the expression.

$\frac{(4 x^{2} + 55 x + 25) (x {(3 x + 5)}^{2})}{x {(3 x + 5)}^{2}} = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.5

Cancel the common factor of $x$ .

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Step 1.5.1

Cancel the common factor.

Step 1.5.2

Rewrite the expression.

$\frac{(4 x^{2} + 55 x + 25) ({(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.6

Cancel the common factor of ${(3 x + 5)}^{2}$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(4 x^{2} + 55 x + 25) {(3 x + 5)}^{2}}{{(3 x + 5)}^{2}} = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.6.2

Divide $4 x^{2} + 55 x + 25$ by $1$ .

$4 x^{2} + 55 x + 25 = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

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.

$4 x^{2} + 55 x + 25 = \frac{A (5 x {(3 x + 5)}^{2})}{x} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.1.2

Divide $A (5 {(3 x + 5)}^{2})$ by $1$ .

$4 x^{2} + 55 x + 25 = A (5 {(3 x + 5)}^{2}) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.2

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 5 A {(3 x + 5)}^{2} + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.3

Rewrite ${(3 x + 5)}^{2}$ as $(3 x + 5) (3 x + 5)$ .

$4 x^{2} + 55 x + 25 = 5 A ((3 x + 5) (3 x + 5)) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.4

Expand $(3 x + 5) (3 x + 5)$ using the FOIL Method.

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Step 1.7.4.1

Apply the distributive property.

$4 x^{2} + 55 x + 25 = 5 A (3 x (3 x + 5) + 5 (3 x + 5)) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.4.2

Apply the distributive property.

$4 x^{2} + 55 x + 25 = 5 A (3 x (3 x) + 3 x \cdot 5 + 5 (3 x + 5)) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.4.3

Apply the distributive property.

$4 x^{2} + 55 x + 25 = 5 A (3 x (3 x) + 3 x \cdot 5 + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5

Simplify and combine like terms.

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Step 1.7.5.1

Simplify each term.

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Step 1.7.5.1.1

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 5 A (3 \cdot (3 x \cdot x) + 3 x \cdot 5 + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.2

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

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Step 1.7.5.1.2.1

Move $x$ .

$4 x^{2} + 55 x + 25 = 5 A (3 \cdot (3 (x \cdot x)) + 3 x \cdot 5 + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.2.2

Multiply $x$ by $x$ .

$4 x^{2} + 55 x + 25 = 5 A (3 \cdot (3 x^{2}) + 3 x \cdot 5 + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.3

Multiply $3$ by $3$ .

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2} + 3 x \cdot 5 + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.4

Multiply $5$ by $3$ .

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2} + 15 x + 5 (3 x) + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.5

Multiply $3$ by $5$ .

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2} + 15 x + 15 x + 5 \cdot 5) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.1.6

Multiply $5$ by $5$ .

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2} + 15 x + 15 x + 25) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.5.2

Add $15 x$ and $15 x$ .

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2} + 30 x + 25) + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.6

Apply the distributive property.

$4 x^{2} + 55 x + 25 = 5 A (9 x^{2}) + 5 A (30 x) + 5 A \cdot 25 + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.7

Simplify.

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Step 1.7.7.1

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 5 \cdot (9 A x^{2}) + 5 A (30 x) + 5 A \cdot 25 + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.7.2

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 5 \cdot (9 A x^{2}) + 5 \cdot (30 A x) + 5 A \cdot 25 + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.7.3

Multiply $25$ by $5$ .

$4 x^{2} + 55 x + 25 = 5 \cdot (9 A x^{2}) + 5 \cdot (30 A x) + 125 A + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.8

Simplify each term.

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Step 1.7.8.1

Multiply $5$ by $9$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 5 \cdot (30 A x) + 125 A + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.8.2

Multiply $5$ by $30$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + \frac{(B) (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.9

Cancel the common factor of ${(3 x + 5)}^{2}$ .

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Step 1.7.9.1

Cancel the common factor.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + \frac{B (5 x {(3 x + 5)}^{2})}{{(3 x + 5)}^{2}} + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.9.2

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

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + (B) (5 x) + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.10

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + \frac{(C) (5 x {(3 x + 5)}^{2})}{3 x + 5}$

Step 1.7.11

Cancel the common factor of ${(3 x + 5)}^{2}$ and $3 x + 5$ .

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Step 1.7.11.1

Factor $3 x + 5$ out of $(C) (5 x {(3 x + 5)}^{2})$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + \frac{(3 x + 5) (C (5 x (3 x + 5)))}{3 x + 5}$

Step 1.7.11.2

Cancel the common factors.

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Step 1.7.11.2.1

Multiply by $1$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + \frac{(3 x + 5) (C (5 x (3 x + 5)))}{(3 x + 5) \cdot 1}$

Step 1.7.11.2.2

Cancel the common factor.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + \frac{(3 x + 5) (C (5 x (3 x + 5)))}{(3 x + 5) \cdot 1}$

Step 1.7.11.2.3

Rewrite the expression.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + \frac{C (5 x (3 x + 5))}{1}$

Step 1.7.11.2.4

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

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + C (5 x (3 x + 5))$

Step 1.7.12

Rewrite using the commutative property of multiplication.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 5 C x (3 x + 5)$

Step 1.7.13

Apply the distributive property.

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 5 C x (3 x) + 5 C x \cdot 5$

Step 1.7.14

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

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Step 1.7.14.1

Move $x$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 5 C (x \cdot x) \cdot 3 + 5 C x \cdot 5$

Step 1.7.14.2

Multiply $x$ by $x$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 5 C x^{2} \cdot 3 + 5 C x \cdot 5$

Step 1.7.15

Multiply $5$ by $5$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 5 C x^{2} \cdot 3 + 25 C x$

Step 1.7.16

Multiply $3$ by $5$ .

$4 x^{2} + 55 x + 25 = 45 A x^{2} + 150 A x + 125 A + 5 B x + 15 C x^{2} + 25 C x$

Step 1.8

Reorder.

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Step 1.8.1

Move $A$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 150 A x + 125 A + 5 B x + 15 C x^{2} + 25 C x$

Step 1.8.2

Move $A$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 150 x A + 125 A + 5 B x + 15 C x^{2} + 25 C x$

Step 1.8.3

Move $B$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 150 x A + 125 A + 5 x B + 15 C x^{2} + 25 C x$

Step 1.8.4

Move $125 A$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 150 x A + 5 x B + 15 C x^{2} + 25 C x + 125 A$

Step 1.8.5

Move $5 x B$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 150 x A + 15 C x^{2} + 5 x B + 25 C x + 125 A$

Step 1.8.6

Move $150 x A$ .

$4 x^{2} + 55 x + 25 = 45 x^{2} A + 15 C x^{2} + 150 x A + 5 x B + 25 C x + 125 A$

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.

$4 = 45 A + 15 C$

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.

$55 = 150 A + 5 B + 25 C$

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.

$25 = 125 A$

Step 2.4

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

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$25 = 125 A$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$25 = 125 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $25 = 125 A$ .

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Step 3.1.1

Rewrite the equation as $125 A = 25$ .

$125 A = 25$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2

Divide each term in $125 A = 25$ by $125$ and simplify.

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Step 3.1.2.1

Divide each term in $125 A = 25$ by $125$ .

$\frac{125 A}{125} = \frac{25}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $125$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{125 A}{125} = \frac{25}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{25}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{25}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{25}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Cancel the common factor of $25$ and $125$ .

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Step 3.1.2.3.1.1

Factor $25$ out of $25$ .

$A = \frac{25 (1)}{125}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.3.1.2

Cancel the common factors.

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Step 3.1.2.3.1.2.1

Factor $25$ out of $125$ .

$A = \frac{25 \cdot 1}{25 \cdot 5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.3.1.2.2

Cancel the common factor.

$A = \frac{25 \cdot 1}{25 \cdot 5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.1.2.3.1.2.3

Rewrite the expression.

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

$A = \frac{1}{5}$

$4 = 45 A + 15 C$

$55 = 150 A + 5 B + 25 C$

Step 3.2

Replace all occurrences of $A$ with $\frac{1}{5}$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $4 = 45 A + 15 C$ with $\frac{1}{5}$ .

$4 = 45 (\frac{1}{5}) + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Cancel the common factor of $5$ .

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Step 3.2.2.1.1

Factor $5$ out of $45$ .

$4 = 5 (9) (\frac{1}{5}) + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

Step 3.2.2.1.2

Cancel the common factor.

$4 = 5 \cdot (9 (\frac{1}{5})) + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

Step 3.2.2.1.3

Rewrite the expression.

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 150 A + 5 B + 25 C$

Step 3.2.3

Replace all occurrences of $A$ in $55 = 150 A + 5 B + 25 C$ with $\frac{1}{5}$ .

$55 = 150 (\frac{1}{5}) + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Cancel the common factor of $5$ .

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Step 3.2.4.1.1

Factor $5$ out of $150$ .

$55 = 5 (30) (\frac{1}{5}) + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

Step 3.2.4.1.2

Cancel the common factor.

$55 = 5 \cdot (30 (\frac{1}{5})) + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

Step 3.2.4.1.3

Rewrite the expression.

$55 = 30 + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 30 + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 30 + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

$55 = 30 + 5 B + 25 C$

$4 = 9 + 15 C$

$A = \frac{1}{5}$

Step 3.3

Solve for $C$ in $4 = 9 + 15 C$ .

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Step 3.3.1

Rewrite the equation as $9 + 15 C = 4$ .

$9 + 15 C = 4$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

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

Subtract $9$ from both sides of the equation.

$15 C = 4 - 9$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.2.2

Subtract $9$ from $4$ .

$15 C = - 5$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$15 C = - 5$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3

Divide each term in $15 C = - 5$ by $15$ and simplify.

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Step 3.3.3.1

Divide each term in $15 C = - 5$ by $15$ .

$\frac{15 C}{15} = \frac{- 5}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $15$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{15 C}{15} = \frac{- 5}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 5}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = \frac{- 5}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = \frac{- 5}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Cancel the common factor of $- 5$ and $15$ .

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Step 3.3.3.3.1.1

Factor $5$ out of $- 5$ .

$C = \frac{5 (- 1)}{15}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.3.1.2

Cancel the common factors.

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Step 3.3.3.3.1.2.1

Factor $5$ out of $15$ .

$C = \frac{5 \cdot - 1}{5 \cdot 3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.3.1.2.2

Cancel the common factor.

$C = \frac{5 \cdot - 1}{5 \cdot 3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.3.1.2.3

Rewrite the expression.

$C = \frac{- 1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = \frac{- 1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = \frac{- 1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.3.3.3.2

Move the negative in front of the fraction.

$C = - \frac{1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = - \frac{1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = - \frac{1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

$C = - \frac{1}{3}$

$55 = 30 + 5 B + 25 C$

$A = \frac{1}{5}$

Step 3.4

Replace all occurrences of $C$ with $- \frac{1}{3}$ in each equation.

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Step 3.4.1

Replace all occurrences of $C$ in $55 = 30 + 5 B + 25 C$ with $- \frac{1}{3}$ .

$55 = 30 + 5 B + 25 (- \frac{1}{3})$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $30 + 5 B + 25 (- \frac{1}{3})$ .

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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 $25 (- \frac{1}{3})$ .

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Step 3.4.2.1.1.1.1

Multiply $- 1$ by $25$ .

$55 = 30 + 5 B - 25 (\frac{1}{3})$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.1.1.2

Combine $- 25$ and $\frac{1}{3}$ .

$55 = 30 + 5 B + \frac{- 25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 30 + 5 B + \frac{- 25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.1.2

Move the negative in front of the fraction.

$55 = 30 + 5 B - \frac{25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 30 + 5 B - \frac{25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.2

To write $30$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$55 = 5 B + 30 \cdot \frac{3}{3} - \frac{25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.3

Combine $30$ and $\frac{3}{3}$ .

$55 = 5 B + \frac{30 \cdot 3}{3} - \frac{25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.4

Combine the numerators over the common denominator.

$55 = 5 B + \frac{30 \cdot 3 - 25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.5

Simplify the numerator.

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Step 3.4.2.1.5.1

Multiply $30$ by $3$ .

$55 = 5 B + \frac{90 - 25}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.4.2.1.5.2

Subtract $25$ from $90$ .

$55 = 5 B + \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 5 B + \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 5 B + \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 5 B + \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$55 = 5 B + \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5

Solve for $B$ in $55 = 5 B + \frac{65}{3}$ .

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Step 3.5.1

Rewrite the equation as $5 B + \frac{65}{3} = 55$ .

$5 B + \frac{65}{3} = 55$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

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

Subtract $\frac{65}{3}$ from both sides of the equation.

$5 B = 55 - \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.2.2

To write $55$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$5 B = 55 \cdot \frac{3}{3} - \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.2.3

Combine $55$ and $\frac{3}{3}$ .

$5 B = \frac{55 \cdot 3}{3} - \frac{65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.2.4

Combine the numerators over the common denominator.

$5 B = \frac{55 \cdot 3 - 65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.2.5

Simplify the numerator.

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Step 3.5.2.5.1

Multiply $55$ by $3$ .

$5 B = \frac{165 - 65}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.2.5.2

Subtract $65$ from $165$ .

$5 B = \frac{100}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$5 B = \frac{100}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$5 B = \frac{100}{3}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3

Divide each term in $5 B = \frac{100}{3}$ by $5$ and simplify.

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Step 3.5.3.1

Divide each term in $5 B = \frac{100}{3}$ by $5$ .

$\frac{5 B}{5} = \frac{\frac{100}{3}}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $5$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{5 B}{5} = \frac{\frac{100}{3}}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{\frac{100}{3}}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$B = \frac{\frac{100}{3}}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

$B = \frac{\frac{100}{3}}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$B = \frac{100}{3} \cdot \frac{1}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.3.2

Cancel the common factor of $5$ .

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Step 3.5.3.3.2.1

Factor $5$ out of $100$ .

$B = \frac{5 (20)}{3} \cdot \frac{1}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.3.2.2

Cancel the common factor.

$B = \frac{5 \cdot 20}{3} \cdot \frac{1}{5}$

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.5.3.3.2.3

Rewrite the expression.

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

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

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

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

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

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

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

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

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

$C = - \frac{1}{3}$

$A = \frac{1}{5}$

Step 3.6

Solve the system of equations.

$B = \frac{20}{3} C = - \frac{1}{3} A = \frac{1}{5}$

Step 3.7

List all of the solutions.

$B = \frac{20}{3}, C = - \frac{1}{3}, A = \frac{1}{5}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{{(3 x + 5)}^{2}} + \frac{C}{3 x + 5}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{5}}{x} + \frac{\frac{20}{3}}{{(3 x + 5)}^{2}} + \frac{- \frac{1}{3}}{3 x + 5}$