Precalculus Examples

$\frac{1}{x (5 x + 3) (4 x^{2} + 1)}$

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 $B$ .

$\frac{A}{x} + \frac{B}{5 x + 3}$

Step 1.2

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A}{x} + \frac{B}{5 x + 3} + \frac{C x + D}{4 x^{2} + 1}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (5 x + 3) (4 x^{2} + 1)$ .

$\frac{1 (x (5 x + 3) (4 x^{2} + 1))}{x (5 x + 3) (4 x^{2} + 1)} = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

Step 1.4.2

Rewrite the expression.

$\frac{1 ((5 x + 3) (4 x^{2} + 1))}{(5 x + 3) (4 x^{2} + 1)} = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.5

Cancel the common factor of $5 x + 3$ .

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Step 1.5.1

Cancel the common factor.

Step 1.5.2

Rewrite the expression.

$\frac{1 (4 x^{2} + 1)}{4 x^{2} + 1} = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.6

Cancel the common factor of $4 x^{2} + 1$ .

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Step 1.6.1

Cancel the common factor.

$\frac{1 (4 x^{2} + 1)}{4 x^{2} + 1} = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.6.2

Rewrite the expression.

$1 = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

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.

$1 = \frac{A (x (5 x + 3) (4 x^{2} + 1))}{x} + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.1.2

Divide $A ((5 x + 3) (4 x^{2} + 1))$ by $1$ .

$1 = A ((5 x + 3) (4 x^{2} + 1)) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.2

Expand $(5 x + 3) (4 x^{2} + 1)$ using the FOIL Method.

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Step 1.7.2.1

Apply the distributive property.

$1 = A (5 x (4 x^{2} + 1) + 3 (4 x^{2} + 1)) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.2.2

Apply the distributive property.

$1 = A (5 x (4 x^{2}) + 5 x \cdot 1 + 3 (4 x^{2} + 1)) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.2.3

Apply the distributive property.

$1 = A (5 x (4 x^{2}) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3

Simplify each term.

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Step 1.7.3.1

Rewrite using the commutative property of multiplication.

$1 = A (5 \cdot (4 x \cdot x^{2}) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.2

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.7.3.2.1

Move $x^{2}$ .

$1 = A (5 \cdot (4 (x^{2} x)) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.2.2

Multiply $x^{2}$ by $x$ .

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Step 1.7.3.2.2.1

Raise $x$ to the power of $1$ .

$1 = A (5 \cdot (4 (x^{2} x)) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1 = A (5 \cdot (4 x^{2 + 1}) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.2.3

Add $2$ and $1$ .

$1 = A (5 \cdot (4 x^{3}) + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.3

Multiply $5$ by $4$ .

$1 = A (20 x^{3} + 5 x \cdot 1 + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.4

Multiply $5$ by $1$ .

$1 = A (20 x^{3} + 5 x + 3 (4 x^{2}) + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.5

Multiply $4$ by $3$ .

$1 = A (20 x^{3} + 5 x + 12 x^{2} + 3 \cdot 1) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.3.6

Multiply $3$ by $1$ .

$1 = A (20 x^{3} + 5 x + 12 x^{2} + 3) + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.4

Apply the distributive property.

$1 = A (20 x^{3}) + A (5 x) + A (12 x^{2}) + A \cdot 3 + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.5

Simplify.

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Step 1.7.5.1

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + A (5 x) + A (12 x^{2}) + A \cdot 3 + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.5.2

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + A (12 x^{2}) + A \cdot 3 + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.5.3

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + A \cdot 3 + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.5.4

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

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 \cdot A + \frac{(B) (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.6

Cancel the common factor of $5 x + 3$ .

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Step 1.7.6.1

Cancel the common factor.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + \frac{B (x (5 x + 3) (4 x^{2} + 1))}{5 x + 3} + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.6.2

Divide $(B) (x (4 x^{2} + 1))$ by $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + (B) (x (4 x^{2} + 1)) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.7

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (x (4 x^{2}) + x \cdot 1) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.8

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 x \cdot x^{2} + x \cdot 1) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.9

Multiply $x$ by $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 x \cdot x^{2} + x) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.10

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.7.10.1

Move $x^{2}$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 (x^{2} x) + x) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.10.2

Multiply $x^{2}$ by $x$ .

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Step 1.7.10.2.1

Raise $x$ to the power of $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 (x^{2} x) + x) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.10.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 x^{2 + 1} + x) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.10.3

Add $2$ and $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 x^{3} + x) + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.11

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + B (4 x^{3}) + B x + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.12

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.13

Cancel the common factor of $4 x^{2} + 1$ .

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Step 1.7.13.1

Cancel the common factor.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + \frac{(C x + D) (x (5 x + 3) (4 x^{2} + 1))}{4 x^{2} + 1}$

Step 1.7.13.2

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

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (x (5 x + 3))$

Step 1.7.14

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (x (5 x) + x \cdot 3)$

Step 1.7.15

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (5 x \cdot x + x \cdot 3)$

Step 1.7.16

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

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (5 x \cdot x + 3 \cdot x)$

Step 1.7.17

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

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Step 1.7.17.1

Move $x$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (5 (x \cdot x) + 3 \cdot x)$

Step 1.7.17.2

Multiply $x$ by $x$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (5 x^{2} + 3 \cdot x)$

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + (C x + D) (5 x^{2} + 3 x)$

Step 1.7.18

Expand $(C x + D) (5 x^{2} + 3 x)$ using the FOIL Method.

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Step 1.7.18.1

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + C x (5 x^{2} + 3 x) + D (5 x^{2} + 3 x)$

Step 1.7.18.2

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + C x (5 x^{2}) + C x (3 x) + D (5 x^{2} + 3 x)$

Step 1.7.18.3

Apply the distributive property.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + C x (5 x^{2}) + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19

Simplify each term.

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Step 1.7.19.1

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x \cdot x^{2} + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.2

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.7.19.2.1

Move $x^{2}$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C (x^{2} x) + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.2.2

Multiply $x^{2}$ by $x$ .

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Step 1.7.19.2.2.1

Raise $x$ to the power of $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C (x^{2} x) + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{2 + 1} + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.2.3

Add $2$ and $1$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + C x (3 x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.3

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x \cdot x + D (5 x^{2}) + D (3 x)$

Step 1.7.19.4

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

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Step 1.7.19.4.1

Move $x$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C (x \cdot x) + D (5 x^{2}) + D (3 x)$

Step 1.7.19.4.2

Multiply $x$ by $x$ .

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + D (5 x^{2}) + D (3 x)$

Step 1.7.19.5

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + D (3 x)$

Step 1.7.19.6

Rewrite using the commutative property of multiplication.

$1 = 20 A x^{3} + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $A$ .

$1 = 20 x^{3} A + 5 A x + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8.2

Move $A$ .

$1 = 20 x^{3} A + 5 x A + 12 A x^{2} + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8.3

Move $A$ .

$1 = 20 x^{3} A + 5 x A + 12 x^{2} A + 3 A + 4 B x^{3} + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8.4

Move $B$ .

$1 = 20 x^{3} A + 5 x A + 12 x^{2} A + 3 A + 4 x^{3} B + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8.5

Reorder $B$ and $x$ .

$1 = 20 x^{3} A + 5 x A + 12 x^{2} A + 3 A + 4 x^{3} B + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x$

Step 1.8.6

Move $3 A$ .

$1 = 20 x^{3} A + 5 x A + 12 x^{2} A + 4 x^{3} B + B x + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 3 D x + 3 A$

Step 1.8.7

Move $B x$ .

$1 = 20 x^{3} A + 5 x A + 12 x^{2} A + 4 x^{3} B + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + B x + 3 D x + 3 A$

Step 1.8.8

Move $5 x A$ .

$1 = 20 x^{3} A + 12 x^{2} A + 4 x^{3} B + 5 C x^{3} + 3 C x^{2} + 5 D x^{2} + 5 x A + B x + 3 D x + 3 A$

Step 1.8.9

Move $12 x^{2} A$ .

$1 = 20 x^{3} A + 4 x^{3} B + 5 C x^{3} + 12 x^{2} A + 3 C x^{2} + 5 D x^{2} + 5 x A + B x + 3 D x + 3 A$

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 $x^{3}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 20 A + 4 B + 5 C$

Step 2.2

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.

$0 = 12 A + 3 C + 5 D$

Step 2.3

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.

$0 = 5 A + B + 3 D$

Step 2.4

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.

$1 = 3 A$

Step 2.5

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

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$1 = 3 A$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$1 = 3 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $1 = 3 A$ .

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Step 3.1.1

Rewrite the equation as $3 A = 1$ .

$3 A = 1$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

Step 3.1.2

Divide each term in $3 A = 1$ by $3$ and simplify.

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Step 3.1.2.1

Divide each term in $3 A = 1$ by $3$ .

$\frac{3 A}{3} = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

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.

$\frac{3 A}{3} = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$A = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$A = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$A = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$A = \frac{1}{3}$

$0 = 20 A + 4 B + 5 C$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $A$ in $0 = 20 A + 4 B + 5 C$ with $\frac{1}{3}$ .

$0 = 20 (\frac{1}{3}) + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Combine $20$ and $\frac{1}{3}$ .

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 12 A + 3 C + 5 D$

$0 = 5 A + B + 3 D$

Step 3.2.3

Replace all occurrences of $A$ in $0 = 12 A + 3 C + 5 D$ with $\frac{1}{3}$ .

$0 = 12 (\frac{1}{3}) + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Cancel the common factor of $3$ .

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Step 3.2.4.1.1

Factor $3$ out of $12$ .

$0 = 3 (4) (\frac{1}{3}) + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

Step 3.2.4.1.2

Cancel the common factor.

$0 = 3 \cdot (4 (\frac{1}{3})) + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

Step 3.2.4.1.3

Rewrite the expression.

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = 5 A + B + 3 D$

Step 3.2.5

Replace all occurrences of $A$ in $0 = 5 A + B + 3 D$ with $\frac{1}{3}$ .

$0 = 5 (\frac{1}{3}) + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

Step 3.2.6

Simplify the right side.

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Step 3.2.6.1

Combine $5$ and $\frac{1}{3}$ .

$0 = \frac{5}{3} + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = \frac{5}{3} + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$0 = \frac{5}{3} + B + 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

Step 3.3

Solve for $B$ in $0 = \frac{5}{3} + B + 3 D$ .

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Step 3.3.1

Rewrite the equation as $\frac{5}{3} + B + 3 D = 0$ .

$\frac{5}{3} + B + 3 D = 0$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $\frac{5}{3}$ from both sides of the equation.

$B + 3 D = - \frac{5}{3}$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

Step 3.3.2.2

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

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$0 = \frac{20}{3} + 4 B + 5 C$

$A = \frac{1}{3}$

Step 3.4

Replace all occurrences of $B$ with $- \frac{5}{3} - 3 D$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $0 = \frac{20}{3} + 4 B + 5 C$ with $- \frac{5}{3} - 3 D$ .

$0 = \frac{20}{3} + 4 (- \frac{5}{3} - 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $\frac{20}{3} + 4 (- \frac{5}{3} - 3 D) + 5 C$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$0 = \frac{20}{3} + 4 (- \frac{5}{3}) + 4 (- 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.1.2

Multiply $4 (- \frac{5}{3})$ .

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Step 3.4.2.1.1.2.1

Multiply $- 1$ by $4$ .

$0 = \frac{20}{3} - 4 (\frac{5}{3}) + 4 (- 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.1.2.2

Combine $- 4$ and $\frac{5}{3}$ .

$0 = \frac{20}{3} + \frac{- 4 \cdot 5}{3} + 4 (- 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.1.2.3

Multiply $- 4$ by $5$ .

$0 = \frac{20}{3} + \frac{- 20}{3} + 4 (- 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = \frac{20}{3} + \frac{- 20}{3} + 4 (- 3 D) + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.1.3

Multiply $- 3$ by $4$ .

$0 = \frac{20}{3} + \frac{- 20}{3} - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.1.4

Move the negative in front of the fraction.

$0 = \frac{20}{3} - \frac{20}{3} - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = \frac{20}{3} - \frac{20}{3} - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Combine the opposite terms in $\frac{20}{3} - \frac{20}{3} - 12 D + 5 C$ .

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Step 3.4.2.1.2.1.1

Combine the numerators over the common denominator.

$0 = \frac{20 - 20}{3} - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.2.1.2

Subtract $20$ from $20$ .

$0 = \frac{0}{3} - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.2.1.3

Divide $0$ by $3$ .

$0 = 0 - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = 0 - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.4.2.1.2.2

Subtract $12 D$ from $0$ .

$0 = - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = - 12 D + 5 C$

$B = - \frac{5}{3} - 3 D$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.5

Reorder $- \frac{5}{3}$ and $- 3 D$ .

$B = - 3 D - \frac{5}{3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6

Solve for $D$ in $B = - 3 D - \frac{5}{3}$ .

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Step 3.6.1

Rewrite the equation as $- 3 D - \frac{5}{3} = B$ .

$- 3 D - \frac{5}{3} = B$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.2

Add $\frac{5}{3}$ to both sides of the equation.

$- 3 D = B + \frac{5}{3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3

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

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Step 3.6.3.1

Divide each term in $- 3 D = B + \frac{5}{3}$ by $- 3$ .

$\frac{- 3 D}{- 3} = \frac{B}{- 3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.2

Simplify the left side.

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Step 3.6.3.2.1

Cancel the common factor of $- 3$ .

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Step 3.6.3.2.1.1

Cancel the common factor.

$\frac{- 3 D}{- 3} = \frac{B}{- 3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.2.1.2

Divide $D$ by $1$ .

$D = \frac{B}{- 3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = \frac{B}{- 3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = \frac{B}{- 3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.3

Simplify the right side.

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Step 3.6.3.3.1

Simplify each term.

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Step 3.6.3.3.1.1

Move the negative in front of the fraction.

$D = - \frac{B}{3} + \frac{\frac{5}{3}}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.3.1.2

Multiply the numerator by the reciprocal of the denominator.

$D = - \frac{B}{3} + \frac{5}{3} \cdot \frac{1}{- 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.3.1.3

Move the negative in front of the fraction.

$D = - \frac{B}{3} + \frac{5}{3} \cdot (- \frac{1}{3})$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.3.1.4

Multiply $\frac{5}{3} (- \frac{1}{3})$ .

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Step 3.6.3.3.1.4.1

Multiply $\frac{5}{3}$ by $\frac{1}{3}$ .

$D = - \frac{B}{3} - \frac{5}{3 \cdot 3}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.6.3.3.1.4.2

Multiply $3$ by $3$ .

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = - 12 D + 5 C$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7

Replace all occurrences of $D$ with $- \frac{B}{3} - \frac{5}{9}$ in each equation.

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Step 3.7.1

Replace all occurrences of $D$ in $0 = - 12 D + 5 C$ with $- \frac{B}{3} - \frac{5}{9}$ .

$0 = - 12 (- \frac{B}{3} - \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2

Simplify the right side.

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Step 3.7.2.1

Simplify each term.

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Step 3.7.2.1.1

Apply the distributive property.

$0 = - 12 (- \frac{B}{3}) - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.2

Cancel the common factor of $3$ .

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Step 3.7.2.1.2.1

Move the leading negative in $- \frac{B}{3}$ into the numerator.

$0 = - 12 \frac{- B}{3} - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.2.2

Factor $3$ out of $- 12$ .

$0 = 3 (- 4) (\frac{- B}{3}) - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.2.3

Cancel the common factor.

$0 = 3 \cdot (- 4 \frac{- B}{3}) - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.2.4

Rewrite the expression.

$0 = - 4 (- B) - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = - 4 (- B) - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.3

Multiply $- 1$ by $- 4$ .

$0 = 4 B - 12 (- \frac{5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.4

Cancel the common factor of $3$ .

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Step 3.7.2.1.4.1

Move the leading negative in $- \frac{5}{9}$ into the numerator.

$0 = 4 B - 12 (\frac{- 5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.4.2

Factor $3$ out of $- 12$ .

$0 = 4 B + 3 (- 4) (\frac{- 5}{9}) + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.4.3

Factor $3$ out of $9$ .

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

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.4.4

Cancel the common factor.

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

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.4.5

Rewrite the expression.

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

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

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

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.5

Combine $- 4$ and $\frac{- 5}{3}$ .

$0 = 4 B + \frac{- 4 \cdot - 5}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.2.1.6

Multiply $- 4$ by $- 5$ .

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$0 = 4 + 3 C + 5 D$

$A = \frac{1}{3}$

Step 3.7.3

Replace all occurrences of $D$ in $0 = 4 + 3 C + 5 D$ with $- \frac{B}{3} - \frac{5}{9}$ .

$0 = 4 + 3 C + 5 (- \frac{B}{3} - \frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4

Simplify the right side.

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Step 3.7.4.1

Simplify $4 + 3 C + 5 (- \frac{B}{3} - \frac{5}{9})$ .

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Step 3.7.4.1.1

Simplify each term.

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Step 3.7.4.1.1.1

Apply the distributive property.

$0 = 4 + 3 C + 5 (- \frac{B}{3}) + 5 (- \frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.2

Multiply $5 (- \frac{B}{3})$ .

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Step 3.7.4.1.1.2.1

Multiply $- 1$ by $5$ .

$0 = 4 + 3 C - 5 \frac{B}{3} + 5 (- \frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.2.2

Combine $- 5$ and $\frac{B}{3}$ .

$0 = 4 + 3 C + \frac{- 5 B}{3} + 5 (- \frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 + 3 C + \frac{- 5 B}{3} + 5 (- \frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.3

Multiply $5 (- \frac{5}{9})$ .

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Step 3.7.4.1.1.3.1

Multiply $- 1$ by $5$ .

$0 = 4 + 3 C + \frac{- 5 B}{3} - 5 (\frac{5}{9})$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.3.2

Combine $- 5$ and $\frac{5}{9}$ .

$0 = 4 + 3 C + \frac{- 5 B}{3} + \frac{- 5 \cdot 5}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.3.3

Multiply $- 5$ by $5$ .

$0 = 4 + 3 C + \frac{- 5 B}{3} + \frac{- 25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 + 3 C + \frac{- 5 B}{3} + \frac{- 25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.4

Simplify each term.

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Step 3.7.4.1.1.4.1

Move the negative in front of the fraction.

$0 = 4 + 3 C - \frac{(5) B}{3} + \frac{- 25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.1.4.2

Move the negative in front of the fraction.

$0 = 4 + 3 C - \frac{5 B}{3} - \frac{25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 + 3 C - \frac{5 B}{3} - \frac{25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 + 3 C - \frac{5 B}{3} - \frac{25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$0 = 3 C - \frac{5 B}{3} + 4 \cdot \frac{9}{9} - \frac{25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.3

Combine $4$ and $\frac{9}{9}$ .

$0 = 3 C - \frac{5 B}{3} + \frac{4 \cdot 9}{9} - \frac{25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.4

Combine the numerators over the common denominator.

$0 = 3 C - \frac{5 B}{3} + \frac{4 \cdot 9 - 25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.5

Simplify the numerator.

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Step 3.7.4.1.5.1

Multiply $4$ by $9$ .

$0 = 3 C - \frac{5 B}{3} + \frac{36 - 25}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.7.4.1.5.2

Subtract $25$ from $36$ .

$0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8

Solve for $C$ in $0 = 3 C - \frac{5 B}{3} + \frac{11}{9}$ .

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Step 3.8.1

Rewrite the equation as $3 C - \frac{5 B}{3} + \frac{11}{9} = 0$ .

$3 C - \frac{5 B}{3} + \frac{11}{9} = 0$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.2

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

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Step 3.8.2.1

Add $\frac{5 B}{3}$ to both sides of the equation.

$3 C + \frac{11}{9} = \frac{5 B}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.2.2

Subtract $\frac{11}{9}$ from both sides of the equation.

$3 C = \frac{5 B}{3} - \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$3 C = \frac{5 B}{3} - \frac{11}{9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3

Divide each term in $3 C = \frac{5 B}{3} - \frac{11}{9}$ by $3$ and simplify.

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Step 3.8.3.1

Divide each term in $3 C = \frac{5 B}{3} - \frac{11}{9}$ by $3$ .

$\frac{3 C}{3} = \frac{\frac{5 B}{3}}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.2

Simplify the left side.

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Step 3.8.3.2.1

Cancel the common factor of $3$ .

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Step 3.8.3.2.1.1

Cancel the common factor.

$\frac{3 C}{3} = \frac{\frac{5 B}{3}}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{\frac{5 B}{3}}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{\frac{5 B}{3}}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{\frac{5 B}{3}}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3

Simplify the right side.

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Step 3.8.3.3.1

Simplify each term.

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Step 3.8.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$C = \frac{5 B}{3} \cdot \frac{1}{3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3.1.2

Multiply $\frac{5 B}{3} \cdot \frac{1}{3}$ .

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Step 3.8.3.3.1.2.1

Multiply $\frac{5 B}{3}$ by $\frac{1}{3}$ .

$C = \frac{5 B}{3 \cdot 3} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3.1.2.2

Multiply $3$ by $3$ .

$C = \frac{5 B}{9} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} + \frac{- \frac{11}{9}}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3.1.3

Multiply the numerator by the reciprocal of the denominator.

$C = \frac{5 B}{9} - \frac{11}{9} \cdot \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3.1.4

Multiply $- \frac{11}{9} \cdot \frac{1}{3}$ .

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Step 3.8.3.3.1.4.1

Multiply $\frac{1}{3}$ by $\frac{11}{9}$ .

$C = \frac{5 B}{9} - \frac{11}{3 \cdot 9}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.8.3.3.1.4.2

Multiply $3$ by $9$ .

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$0 = 4 B + \frac{20}{3} + 5 C$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9

Replace all occurrences of $C$ with $\frac{5 B}{9} - \frac{11}{27}$ in each equation.

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Step 3.9.1

Replace all occurrences of $C$ in $0 = 4 B + \frac{20}{3} + 5 C$ with $\frac{5 B}{9} - \frac{11}{27}$ .

$0 = 4 B + \frac{20}{3} + 5 (\frac{5 B}{9} - \frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2

Simplify the right side.

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Step 3.9.2.1

Simplify $4 B + \frac{20}{3} + 5 (\frac{5 B}{9} - \frac{11}{27})$ .

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Step 3.9.2.1.1

Simplify each term.

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Step 3.9.2.1.1.1

Apply the distributive property.

$0 = 4 B + \frac{20}{3} + 5 (\frac{5 B}{9}) + 5 (- \frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.2

Multiply $5 \frac{5 B}{9}$ .

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Step 3.9.2.1.1.2.1

Combine $5$ and $\frac{5 B}{9}$ .

$0 = 4 B + \frac{20}{3} + \frac{5 (5 B)}{9} + 5 (- \frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.2.2

Multiply $5$ by $5$ .

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} + 5 (- \frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} + 5 (- \frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.3

Multiply $5 (- \frac{11}{27})$ .

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Step 3.9.2.1.1.3.1

Multiply $- 1$ by $5$ .

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} - 5 (\frac{11}{27})$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.3.2

Combine $- 5$ and $\frac{11}{27}$ .

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} + \frac{- 5 \cdot 11}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.3.3

Multiply $- 5$ by $11$ .

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} + \frac{- 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} + \frac{- 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.1.4

Move the negative in front of the fraction.

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = 4 B + \frac{20}{3} + \frac{25 B}{9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.2

To write $4 B$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$0 = 4 B \cdot \frac{9}{9} + \frac{25 B}{9} + \frac{20}{3} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.3

Combine $4 B$ and $\frac{9}{9}$ .

$0 = \frac{4 B \cdot 9}{9} + \frac{25 B}{9} + \frac{20}{3} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.4

Combine the numerators over the common denominator.

$0 = \frac{4 B \cdot 9 + 25 B}{9} + \frac{20}{3} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5

Find the common denominator.

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Step 3.9.2.1.5.1

Multiply $\frac{4 B \cdot 9 + 25 B}{9}$ by $\frac{3}{3}$ .

$0 = \frac{4 B \cdot 9 + 25 B}{9} \cdot \frac{3}{3} + \frac{20}{3} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.2

Multiply $\frac{4 B \cdot 9 + 25 B}{9}$ by $\frac{3}{3}$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{9 \cdot 3} + \frac{20}{3} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.3

Multiply $\frac{20}{3}$ by $\frac{9}{9}$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{9 \cdot 3} + \frac{20}{3} \cdot \frac{9}{9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.4

Multiply $\frac{20}{3}$ by $\frac{9}{9}$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{9 \cdot 3} + \frac{20 \cdot 9}{3 \cdot 9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.5

Reorder the factors of $9 \cdot 3$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{3 \cdot 9} + \frac{20 \cdot 9}{3 \cdot 9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.6

Multiply $3$ by $9$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{27} + \frac{20 \cdot 9}{3 \cdot 9} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.5.7

Multiply $3$ by $9$ .

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{27} + \frac{20 \cdot 9}{27} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3}{27} + \frac{20 \cdot 9}{27} - \frac{55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.6

Combine the numerators over the common denominator.

$0 = \frac{(4 B \cdot 9 + 25 B) \cdot 3 + 20 \cdot 9 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.7

Simplify each term.

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Step 3.9.2.1.7.1

Multiply $9$ by $4$ .

$0 = \frac{(36 B + 25 B) \cdot 3 + 20 \cdot 9 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.7.2

Add $36 B$ and $25 B$ .

$0 = \frac{61 B \cdot 3 + 20 \cdot 9 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.7.3

Multiply $3$ by $61$ .

$0 = \frac{183 B + 20 \cdot 9 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.7.4

Multiply $20$ by $9$ .

$0 = \frac{183 B + 180 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = \frac{183 B + 180 - 55}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.9.2.1.8

Subtract $55$ from $180$ .

$0 = \frac{183 B + 125}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = \frac{183 B + 125}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = \frac{183 B + 125}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$0 = \frac{183 B + 125}{27}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10

Solve for $B$ in $0 = \frac{183 B + 125}{27}$ .

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Step 3.10.1

Set the numerator equal to zero.

$183 B + 125 = 0$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10.2

Solve the equation for $B$ .

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Step 3.10.2.1

Subtract $125$ from both sides of the equation.

$183 B = - 125$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10.2.2

Divide each term in $183 B = - 125$ by $183$ and simplify.

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Step 3.10.2.2.1

Divide each term in $183 B = - 125$ by $183$ .

$\frac{183 B}{183} = \frac{- 125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10.2.2.2

Simplify the left side.

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Step 3.10.2.2.2.1

Cancel the common factor of $183$ .

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Step 3.10.2.2.2.1.1

Cancel the common factor.

$\frac{183 B}{183} = \frac{- 125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10.2.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = \frac{- 125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = \frac{- 125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.10.2.2.3

Simplify the right side.

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Step 3.10.2.2.3.1

Move the negative in front of the fraction.

$B = - \frac{125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = - \frac{125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = - \frac{125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = - \frac{125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$B = - \frac{125}{183}$

$C = \frac{5 B}{9} - \frac{11}{27}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11

Replace all occurrences of $B$ with $- \frac{125}{183}$ in each equation.

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Step 3.11.1

Replace all occurrences of $B$ in $C = \frac{5 B}{9} - \frac{11}{27}$ with $- \frac{125}{183}$ .

$C = \frac{5 (- \frac{125}{183})}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2

Simplify the right side.

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Step 3.11.2.1

Simplify $\frac{5 (- \frac{125}{183})}{9} - \frac{11}{27}$ .

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Step 3.11.2.1.1

Simplify each term.

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Step 3.11.2.1.1.1

Simplify the numerator.

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Step 3.11.2.1.1.1.1

Multiply $- 1$ by $5$ .

$C = \frac{- 5 (\frac{125}{183})}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.1.2

Combine $- 5$ and $\frac{125}{183}$ .

$C = \frac{\frac{- 5 \cdot 125}{183}}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{\frac{- 5 \cdot 125}{183}}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.2

Multiply $- 5$ by $125$ .

$C = \frac{\frac{- 625}{183}}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.3

Move the negative in front of the fraction.

$C = \frac{- \frac{625}{183}}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.4

Multiply the numerator by the reciprocal of the denominator.

$C = - \frac{625}{183} \cdot \frac{1}{9} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.5

Multiply $- \frac{625}{183} \cdot \frac{1}{9}$ .

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Step 3.11.2.1.1.5.1

Multiply $\frac{1}{9}$ by $\frac{625}{183}$ .

$C = - \frac{625}{9 \cdot 183} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.1.5.2

Multiply $9$ by $183$ .

$C = - \frac{625}{1647} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = - \frac{625}{1647} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = - \frac{625}{1647} - \frac{11}{27}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.2

To write $- \frac{11}{27}$ as a fraction with a common denominator, multiply by $\frac{61}{61}$ .

$C = - \frac{625}{1647} - \frac{11}{27} \cdot \frac{61}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.3

Write each expression with a common denominator of $1647$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.11.2.1.3.1

Multiply $\frac{11}{27}$ by $\frac{61}{61}$ .

$C = - \frac{625}{1647} - \frac{11 \cdot 61}{27 \cdot 61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.3.2

Multiply $27$ by $61$ .

$C = - \frac{625}{1647} - \frac{11 \cdot 61}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = - \frac{625}{1647} - \frac{11 \cdot 61}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.4

Combine the numerators over the common denominator.

$C = \frac{- 625 - 11 \cdot 61}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.5

Simplify the numerator.

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Step 3.11.2.1.5.1

Multiply $- 11$ by $61$ .

$C = \frac{- 625 - 671}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.5.2

Subtract $671$ from $- 625$ .

$C = \frac{- 1296}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{- 1296}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.6

Cancel the common factor of $- 1296$ and $1647$ .

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Step 3.11.2.1.6.1

Factor $27$ out of $- 1296$ .

$C = \frac{27 (- 48)}{1647}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.6.2

Cancel the common factors.

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Step 3.11.2.1.6.2.1

Factor $27$ out of $1647$ .

$C = \frac{27 \cdot - 48}{27 \cdot 61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.6.2.2

Cancel the common factor.

$C = \frac{27 \cdot - 48}{27 \cdot 61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.6.2.3

Rewrite the expression.

$C = \frac{- 48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{- 48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = \frac{- 48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.2.1.7

Move the negative in front of the fraction.

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$D = - \frac{B}{3} - \frac{5}{9}$

$A = \frac{1}{3}$

Step 3.11.3

Replace all occurrences of $B$ in $D = - \frac{B}{3} - \frac{5}{9}$ with $- \frac{125}{183}$ .

$D = - \frac{- \frac{125}{183}}{3} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4

Simplify the right side.

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Step 3.11.4.1

Simplify $- \frac{- \frac{125}{183}}{3} - \frac{5}{9}$ .

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Step 3.11.4.1.1

Simplify each term.

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Step 3.11.4.1.1.1

Multiply the numerator by the reciprocal of the denominator.

$D = - (- \frac{125}{183} \cdot \frac{1}{3}) - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.1.2

Multiply $- \frac{125}{183} \cdot \frac{1}{3}$ .

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Step 3.11.4.1.1.2.1

Multiply $\frac{1}{3}$ by $\frac{125}{183}$ .

$D = \frac{125}{3 \cdot 183} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.1.2.2

Multiply $3$ by $183$ .

$D = \frac{125}{549} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{125}{549} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.1.3

Multiply $- - \frac{125}{549}$ .

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Step 3.11.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$D = 1 (\frac{125}{549}) - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.1.3.2

Multiply $\frac{125}{549}$ by $1$ .

$D = \frac{125}{549} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{125}{549} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{125}{549} - \frac{5}{9}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.2

To write $- \frac{5}{9}$ as a fraction with a common denominator, multiply by $\frac{61}{61}$ .

$D = \frac{125}{549} - \frac{5}{9} \cdot \frac{61}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.3

Write each expression with a common denominator of $549$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.11.4.1.3.1

Multiply $\frac{5}{9}$ by $\frac{61}{61}$ .

$D = \frac{125}{549} - \frac{5 \cdot 61}{9 \cdot 61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.3.2

Multiply $9$ by $61$ .

$D = \frac{125}{549} - \frac{5 \cdot 61}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{125}{549} - \frac{5 \cdot 61}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.4

Combine the numerators over the common denominator.

$D = \frac{125 - 5 \cdot 61}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.5

Simplify the numerator.

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Step 3.11.4.1.5.1

Multiply $- 5$ by $61$ .

$D = \frac{125 - 305}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.5.2

Subtract $305$ from $125$ .

$D = \frac{- 180}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{- 180}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.6

Cancel the common factor of $- 180$ and $549$ .

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Step 3.11.4.1.6.1

Factor $9$ out of $- 180$ .

$D = \frac{9 (- 20)}{549}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.6.2

Cancel the common factors.

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Step 3.11.4.1.6.2.1

Factor $9$ out of $549$ .

$D = \frac{9 \cdot - 20}{9 \cdot 61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.6.2.2

Cancel the common factor.

$D = \frac{9 \cdot - 20}{9 \cdot 61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.6.2.3

Rewrite the expression.

$D = \frac{- 20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{- 20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = \frac{- 20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.11.4.1.7

Move the negative in front of the fraction.

$D = - \frac{20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = - \frac{20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = - \frac{20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

$D = - \frac{20}{61}$

$C = - \frac{48}{61}$

$B = - \frac{125}{183}$

$A = \frac{1}{3}$

Step 3.12

List all of the solutions.

$D = - \frac{20}{61}, C = - \frac{48}{61}, B = - \frac{125}{183}, A = \frac{1}{3}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{5 x + 3} + \frac{C x + D}{4 x^{2} + 1}$ with the values found for $A$ , $B$ , $C$ , and $D$ .

$\frac{\frac{1}{3}}{x} + \frac{- \frac{125}{183}}{5 x + 3} + \frac{- \frac{48}{61 x} - \frac{20}{61}}{4 x^{2} + 1}$

Step 5

Simplify each term.

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Step 5.1

Combine $x$ and $\frac{48}{61}$ .

$\frac{\frac{1}{3}}{x} + \frac{- \frac{125}{183}}{5 x + 3} + \frac{- \frac{x \cdot 48}{61} - \frac{20}{61}}{4 x^{2} + 1}$

Step 5.2

Move $48$ to the left of $x$ .

$\frac{\frac{1}{3}}{x} + \frac{- \frac{125}{183}}{5 x + 3} + \frac{- \frac{48 x}{61} - \frac{20}{61}}{4 x^{2} + 1}$