Precalculus Examples

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

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 $B$ .

$\frac{A}{x} + \frac{B}{2 x + 5}$

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.

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

Step 1.3

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

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

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.

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

Step 1.5

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

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Step 1.5.1

Cancel the common factor.

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Cancel the common factor of $3 x^{2} + 1$ .

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Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

Rewrite the expression.

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

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.

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

Step 1.7.1.2

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

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

Step 1.7.2

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

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Step 1.7.2.1

Apply the distributive property.

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

Step 1.7.2.2

Apply the distributive property.

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

Step 1.7.2.3

Apply the distributive property.

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

Step 1.7.3

Simplify each term.

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Step 1.7.3.1

Rewrite using the commutative property of multiplication.

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

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 $x^{2}$ .

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

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 $1$ .

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

Step 1.7.3.2.2.2

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

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

Step 1.7.3.2.3

Add $2$ and $1$ .

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

Step 1.7.3.3

Multiply $2$ by $3$ .

$1 = A (6 x^{3} + 2 x \cdot 1 + 5 (3 x^{2}) + 5 \cdot 1) + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.3.4

Multiply $2$ by $1$ .

$1 = A (6 x^{3} + 2 x + 5 (3 x^{2}) + 5 \cdot 1) + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.3.5

Multiply $3$ by $5$ .

$1 = A (6 x^{3} + 2 x + 15 x^{2} + 5 \cdot 1) + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.3.6

Multiply $5$ by $1$ .

$1 = A (6 x^{3} + 2 x + 15 x^{2} + 5) + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.4

Apply the distributive property.

$1 = A (6 x^{3}) + A (2 x) + A (15 x^{2}) + A \cdot 5 + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.5

Simplify.

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Step 1.7.5.1

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + A (2 x) + A (15 x^{2}) + A \cdot 5 + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.5.2

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + A (15 x^{2}) + A \cdot 5 + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.5.3

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + A \cdot 5 + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.5.4

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

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 \cdot A + \frac{(B) (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.6

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

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Step 1.7.6.1

Cancel the common factor.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + \frac{B (x (2 x + 5) (3 x^{2} + 1))}{2 x + 5} + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.6.2

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

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + (B) (x (3 x^{2} + 1)) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.7

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (x (3 x^{2}) + x \cdot 1) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.8

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 x \cdot x^{2} + x \cdot 1) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.9

Multiply $x$ by $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 x \cdot x^{2} + x) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.10

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

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Step 1.7.10.1

Move $x^{2}$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 (x^{2} x) + x) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

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 $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 (x^{2} x) + x) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.10.2.2

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

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 x^{2 + 1} + x) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.10.3

Add $2$ and $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 x^{3} + x) + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.11

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + B (3 x^{3}) + B x + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.12

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.13

Cancel the common factor of $3 x^{2} + 1$ .

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Step 1.7.13.1

Cancel the common factor.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + \frac{(C x + D) (x (2 x + 5) (3 x^{2} + 1))}{3 x^{2} + 1}$

Step 1.7.13.2

Divide $(C x + D) (x (2 x + 5))$ by $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (x (2 x + 5))$

Step 1.7.14

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (x (2 x) + x \cdot 5)$

Step 1.7.15

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (2 x \cdot x + x \cdot 5)$

Step 1.7.16

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

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (2 x \cdot x + 5 \cdot x)$

Step 1.7.17

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

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Step 1.7.17.1

Move $x$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (2 (x \cdot x) + 5 \cdot x)$

Step 1.7.17.2

Multiply $x$ by $x$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (2 x^{2} + 5 \cdot x)$

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + (C x + D) (2 x^{2} + 5 x)$

Step 1.7.18

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

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Step 1.7.18.1

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + C x (2 x^{2} + 5 x) + D (2 x^{2} + 5 x)$

Step 1.7.18.2

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + C x (2 x^{2}) + C x (5 x) + D (2 x^{2} + 5 x)$

Step 1.7.18.3

Apply the distributive property.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + C x (2 x^{2}) + C x (5 x) + D (2 x^{2}) + D (5 x)$

Step 1.7.19

Simplify each term.

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Step 1.7.19.1

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x \cdot x^{2} + C x (5 x) + D (2 x^{2}) + D (5 x)$

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 $x^{2}$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C (x^{2} x) + C x (5 x) + D (2 x^{2}) + D (5 x)$

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 $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C (x^{2} x) + C x (5 x) + D (2 x^{2}) + D (5 x)$

Step 1.7.19.2.2.2

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

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{2 + 1} + C x (5 x) + D (2 x^{2}) + D (5 x)$

Step 1.7.19.2.3

Add $2$ and $1$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + C x (5 x) + D (2 x^{2}) + D (5 x)$

Step 1.7.19.3

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x \cdot x + D (2 x^{2}) + D (5 x)$

Step 1.7.19.4

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

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Step 1.7.19.4.1

Move $x$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C (x \cdot x) + D (2 x^{2}) + D (5 x)$

Step 1.7.19.4.2

Multiply $x$ by $x$ .

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + D (2 x^{2}) + D (5 x)$

Step 1.7.19.5

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + D (5 x)$

Step 1.7.19.6

Rewrite using the commutative property of multiplication.

$1 = 6 A x^{3} + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $A$ .

$1 = 6 x^{3} A + 2 A x + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8.2

Move $A$ .

$1 = 6 x^{3} A + 2 x A + 15 A x^{2} + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8.3

Move $A$ .

$1 = 6 x^{3} A + 2 x A + 15 x^{2} A + 5 A + 3 B x^{3} + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8.4

Move $B$ .

$1 = 6 x^{3} A + 2 x A + 15 x^{2} A + 5 A + 3 x^{3} B + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8.5

Reorder $B$ and $x$ .

$1 = 6 x^{3} A + 2 x A + 15 x^{2} A + 5 A + 3 x^{3} B + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x$

Step 1.8.6

Move $5 A$ .

$1 = 6 x^{3} A + 2 x A + 15 x^{2} A + 3 x^{3} B + B x + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 5 D x + 5 A$

Step 1.8.7

Move $B x$ .

$1 = 6 x^{3} A + 2 x A + 15 x^{2} A + 3 x^{3} B + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + B x + 5 D x + 5 A$

Step 1.8.8

Move $2 x A$ .

$1 = 6 x^{3} A + 15 x^{2} A + 3 x^{3} B + 2 C x^{3} + 5 C x^{2} + 2 D x^{2} + 2 x A + B x + 5 D x + 5 A$

Step 1.8.9

Move $15 x^{2} A$ .

$1 = 6 x^{3} A + 3 x^{3} B + 2 C x^{3} + 15 x^{2} A + 5 C x^{2} + 2 D x^{2} + 2 x A + B x + 5 D x + 5 A$

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

$0 = 6 A + 3 B + 2 C$

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.

$0 = 15 A + 5 C + 2 D$

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.

$0 = 2 A + B + 5 D$

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.

$1 = 5 A$

Step 2.5

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

$1 = 5 A$

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

$1 = 5 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $1 = 5 A$ .

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Step 3.1.1

Rewrite the equation as $5 A = 1$ .

$5 A = 1$

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

Step 3.1.2

Divide each term in $5 A = 1$ by $5$ and simplify.

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Step 3.1.2.1

Divide each term in $5 A = 1$ by $5$ .

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $5$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

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

$0 = 6 A + 3 B + 2 C$

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

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 $0 = 6 A + 3 B + 2 C$ with $\frac{1}{5}$ .

$0 = 6 (\frac{1}{5}) + 3 B + 2 C$

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

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Combine $6$ and $\frac{1}{5}$ .

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 15 A + 5 C + 2 D$

$0 = 2 A + B + 5 D$

Step 3.2.3

Replace all occurrences of $A$ in $0 = 15 A + 5 C + 2 D$ with $\frac{1}{5}$ .

$0 = 15 (\frac{1}{5}) + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

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 $15$ .

$0 = 5 (3) (\frac{1}{5}) + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

Step 3.2.4.1.2

Cancel the common factor.

$0 = 5 \cdot (3 (\frac{1}{5})) + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

Step 3.2.4.1.3

Rewrite the expression.

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = 2 A + B + 5 D$

Step 3.2.5

Replace all occurrences of $A$ in $0 = 2 A + B + 5 D$ with $\frac{1}{5}$ .

$0 = 2 (\frac{1}{5}) + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

Step 3.2.6

Simplify the right side.

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Step 3.2.6.1

Combine $2$ and $\frac{1}{5}$ .

$0 = \frac{2}{5} + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = \frac{2}{5} + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$0 = \frac{2}{5} + B + 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

Step 3.3

Solve for $B$ in $0 = \frac{2}{5} + B + 5 D$ .

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Step 3.3.1

Rewrite the equation as $\frac{2}{5} + B + 5 D = 0$ .

$\frac{2}{5} + B + 5 D = 0$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

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 $\frac{2}{5}$ from both sides of the equation.

$B + 5 D = - \frac{2}{5}$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

Step 3.3.2.2

Subtract $5 D$ from both sides of the equation.

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

$0 = \frac{6}{5} + 3 B + 2 C$

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

Step 3.4

Replace all occurrences of $B$ with $- \frac{2}{5} - 5 D$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $0 = \frac{6}{5} + 3 B + 2 C$ with $- \frac{2}{5} - 5 D$ .

$0 = \frac{6}{5} + 3 (- \frac{2}{5} - 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $\frac{6}{5} + 3 (- \frac{2}{5} - 5 D) + 2 C$ .

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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.

$0 = \frac{6}{5} + 3 (- \frac{2}{5}) + 3 (- 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.1.2

Multiply $3 (- \frac{2}{5})$ .

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Step 3.4.2.1.1.2.1

Multiply $- 1$ by $3$ .

$0 = \frac{6}{5} - 3 (\frac{2}{5}) + 3 (- 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.1.2.2

Combine $- 3$ and $\frac{2}{5}$ .

$0 = \frac{6}{5} + \frac{- 3 \cdot 2}{5} + 3 (- 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.1.2.3

Multiply $- 3$ by $2$ .

$0 = \frac{6}{5} + \frac{- 6}{5} + 3 (- 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = \frac{6}{5} + \frac{- 6}{5} + 3 (- 5 D) + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.1.3

Multiply $- 5$ by $3$ .

$0 = \frac{6}{5} + \frac{- 6}{5} - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.1.4

Move the negative in front of the fraction.

$0 = \frac{6}{5} - \frac{6}{5} - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = \frac{6}{5} - \frac{6}{5} - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

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 $\frac{6}{5} - \frac{6}{5} - 15 D + 2 C$ .

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Step 3.4.2.1.2.1.1

Combine the numerators over the common denominator.

$0 = \frac{6 - 6}{5} - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.2.1.2

Subtract $6$ from $6$ .

$0 = \frac{0}{5} - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.2.1.3

Divide $0$ by $5$ .

$0 = 0 - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = 0 - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.4.2.1.2.2

Subtract $15 D$ from $0$ .

$0 = - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

$0 = - 15 D + 2 C$

$B = - \frac{2}{5} - 5 D$

$0 = 3 + 5 C + 2 D$

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

Step 3.5

Reorder $- \frac{2}{5}$ and $- 5 D$ .

$B = - 5 D - \frac{2}{5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6

Solve for $D$ in $B = - 5 D - \frac{2}{5}$ .

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Step 3.6.1

Rewrite the equation as $- 5 D - \frac{2}{5} = B$ .

$- 5 D - \frac{2}{5} = B$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.2

Add $\frac{2}{5}$ to both sides of the equation.

$- 5 D = B + \frac{2}{5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3

Divide each term in $- 5 D = B + \frac{2}{5}$ by $- 5$ and simplify.

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Step 3.6.3.1

Divide each term in $- 5 D = B + \frac{2}{5}$ by $- 5$ .

$\frac{- 5 D}{- 5} = \frac{B}{- 5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.2

Simplify the left side.

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Step 3.6.3.2.1

Cancel the common factor of $- 5$ .

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Step 3.6.3.2.1.1

Cancel the common factor.

$\frac{- 5 D}{- 5} = \frac{B}{- 5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.2.1.2

Divide $D$ by $1$ .

$D = \frac{B}{- 5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = \frac{B}{- 5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = \frac{B}{- 5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

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.

$D = - \frac{B}{5} + \frac{\frac{2}{5}}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.3.1.2

Multiply the numerator by the reciprocal of the denominator.

$D = - \frac{B}{5} + \frac{2}{5} \cdot \frac{1}{- 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.3.1.3

Move the negative in front of the fraction.

$D = - \frac{B}{5} + \frac{2}{5} \cdot (- \frac{1}{5})$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.3.1.4

Multiply $\frac{2}{5} (- \frac{1}{5})$ .

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Step 3.6.3.3.1.4.1

Multiply $\frac{2}{5}$ by $\frac{1}{5}$ .

$D = - \frac{B}{5} - \frac{2}{5 \cdot 5}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.6.3.3.1.4.2

Multiply $5$ by $5$ .

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = - 15 D + 2 C$

$0 = 3 + 5 C + 2 D$

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

Step 3.7

Replace all occurrences of $D$ with $- \frac{B}{5} - \frac{2}{25}$ in each equation.

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Step 3.7.1

Replace all occurrences of $D$ in $0 = - 15 D + 2 C$ with $- \frac{B}{5} - \frac{2}{25}$ .

$0 = - 15 (- \frac{B}{5} - \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

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.

$0 = - 15 (- \frac{B}{5}) - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.2

Cancel the common factor of $5$ .

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Step 3.7.2.1.2.1

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

$0 = - 15 \frac{- B}{5} - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.2.2

Factor $5$ out of $- 15$ .

$0 = 5 (- 3) (\frac{- B}{5}) - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.2.3

Cancel the common factor.

$0 = 5 \cdot (- 3 \frac{- B}{5}) - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.2.4

Rewrite the expression.

$0 = - 3 (- B) - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

$0 = - 3 (- B) - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.3

Multiply $- 1$ by $- 3$ .

$0 = 3 B - 15 (- \frac{2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.4

Cancel the common factor of $5$ .

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Step 3.7.2.1.4.1

Move the leading negative in $- \frac{2}{25}$ into the numerator.

$0 = 3 B - 15 (\frac{- 2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.4.2

Factor $5$ out of $- 15$ .

$0 = 3 B + 5 (- 3) (\frac{- 2}{25}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.4.3

Factor $5$ out of $25$ .

$0 = 3 B + 5 \cdot (- 3 \frac{- 2}{5 \cdot 5}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.4.4

Cancel the common factor.

$0 = 3 B + 5 \cdot (- 3 \frac{- 2}{5 \cdot 5}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.4.5

Rewrite the expression.

$0 = 3 B - 3 (\frac{- 2}{5}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

$0 = 3 B - 3 (\frac{- 2}{5}) + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.5

Combine $- 3$ and $\frac{- 2}{5}$ .

$0 = 3 B + \frac{- 3 \cdot - 2}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.2.1.6

Multiply $- 3$ by $- 2$ .

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

$0 = 3 + 5 C + 2 D$

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

Step 3.7.3

Replace all occurrences of $D$ in $0 = 3 + 5 C + 2 D$ with $- \frac{B}{5} - \frac{2}{25}$ .

$0 = 3 + 5 C + 2 (- \frac{B}{5} - \frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4

Simplify the right side.

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Step 3.7.4.1

Simplify $3 + 5 C + 2 (- \frac{B}{5} - \frac{2}{25})$ .

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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.

$0 = 3 + 5 C + 2 (- \frac{B}{5}) + 2 (- \frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.2

Multiply $2 (- \frac{B}{5})$ .

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Step 3.7.4.1.1.2.1

Multiply $- 1$ by $2$ .

$0 = 3 + 5 C - 2 \frac{B}{5} + 2 (- \frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.2.2

Combine $- 2$ and $\frac{B}{5}$ .

$0 = 3 + 5 C + \frac{- 2 B}{5} + 2 (- \frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 + 5 C + \frac{- 2 B}{5} + 2 (- \frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.3

Multiply $2 (- \frac{2}{25})$ .

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Step 3.7.4.1.1.3.1

Multiply $- 1$ by $2$ .

$0 = 3 + 5 C + \frac{- 2 B}{5} - 2 (\frac{2}{25})$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.3.2

Combine $- 2$ and $\frac{2}{25}$ .

$0 = 3 + 5 C + \frac{- 2 B}{5} + \frac{- 2 \cdot 2}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.3.3

Multiply $- 2$ by $2$ .

$0 = 3 + 5 C + \frac{- 2 B}{5} + \frac{- 4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 + 5 C + \frac{- 2 B}{5} + \frac{- 4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

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.

$0 = 3 + 5 C - \frac{(2) B}{5} + \frac{- 4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.1.4.2

Move the negative in front of the fraction.

$0 = 3 + 5 C - \frac{2 B}{5} - \frac{4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 + 5 C - \frac{2 B}{5} - \frac{4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 + 5 C - \frac{2 B}{5} - \frac{4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$0 = 5 C - \frac{2 B}{5} + 3 \cdot \frac{25}{25} - \frac{4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.3

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

$0 = 5 C - \frac{2 B}{5} + \frac{3 \cdot 25}{25} - \frac{4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.4

Combine the numerators over the common denominator.

$0 = 5 C - \frac{2 B}{5} + \frac{3 \cdot 25 - 4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.5

Simplify the numerator.

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Step 3.7.4.1.5.1

Multiply $3$ by $25$ .

$0 = 5 C - \frac{2 B}{5} + \frac{75 - 4}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.7.4.1.5.2

Subtract $4$ from $75$ .

$0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8

Solve for $C$ in $0 = 5 C - \frac{2 B}{5} + \frac{71}{25}$ .

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Step 3.8.1

Rewrite the equation as $5 C - \frac{2 B}{5} + \frac{71}{25} = 0$ .

$5 C - \frac{2 B}{5} + \frac{71}{25} = 0$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

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 $\frac{2 B}{5}$ to both sides of the equation.

$5 C + \frac{71}{25} = \frac{2 B}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.2.2

Subtract $\frac{71}{25}$ from both sides of the equation.

$5 C = \frac{2 B}{5} - \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$5 C = \frac{2 B}{5} - \frac{71}{25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3

Divide each term in $5 C = \frac{2 B}{5} - \frac{71}{25}$ by $5$ and simplify.

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Step 3.8.3.1

Divide each term in $5 C = \frac{2 B}{5} - \frac{71}{25}$ by $5$ .

$\frac{5 C}{5} = \frac{\frac{2 B}{5}}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.2

Simplify the left side.

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Step 3.8.3.2.1

Cancel the common factor of $5$ .

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Step 3.8.3.2.1.1

Cancel the common factor.

$\frac{5 C}{5} = \frac{\frac{2 B}{5}}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{\frac{2 B}{5}}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{\frac{2 B}{5}}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{\frac{2 B}{5}}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

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.

$C = \frac{2 B}{5} \cdot \frac{1}{5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.3.1.2

Multiply $\frac{2 B}{5} \cdot \frac{1}{5}$ .

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Step 3.8.3.3.1.2.1

Multiply $\frac{2 B}{5}$ by $\frac{1}{5}$ .

$C = \frac{2 B}{5 \cdot 5} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.3.1.2.2

Multiply $5$ by $5$ .

$C = \frac{2 B}{25} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} + \frac{- \frac{71}{25}}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.3.1.3

Multiply the numerator by the reciprocal of the denominator.

$C = \frac{2 B}{25} - \frac{71}{25} \cdot \frac{1}{5}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.3.1.4

Multiply $- \frac{71}{25} \cdot \frac{1}{5}$ .

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Step 3.8.3.3.1.4.1

Multiply $\frac{1}{5}$ by $\frac{71}{25}$ .

$C = \frac{2 B}{25} - \frac{71}{5 \cdot 25}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.8.3.3.1.4.2

Multiply $5$ by $25$ .

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{2 B}{25} - \frac{71}{125}$

$0 = 3 B + \frac{6}{5} + 2 C$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9

Replace all occurrences of $C$ with $\frac{2 B}{25} - \frac{71}{125}$ in each equation.

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Step 3.9.1

Replace all occurrences of $C$ in $0 = 3 B + \frac{6}{5} + 2 C$ with $\frac{2 B}{25} - \frac{71}{125}$ .

$0 = 3 B + \frac{6}{5} + 2 (\frac{2 B}{25} - \frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2

Simplify the right side.

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Step 3.9.2.1

Simplify $3 B + \frac{6}{5} + 2 (\frac{2 B}{25} - \frac{71}{125})$ .

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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.

$0 = 3 B + \frac{6}{5} + 2 (\frac{2 B}{25}) + 2 (- \frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.2

Multiply $2 \frac{2 B}{25}$ .

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Step 3.9.2.1.1.2.1

Combine $2$ and $\frac{2 B}{25}$ .

$0 = 3 B + \frac{6}{5} + \frac{2 (2 B)}{25} + 2 (- \frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.2.2

Multiply $2$ by $2$ .

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} + 2 (- \frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} + 2 (- \frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.3

Multiply $2 (- \frac{71}{125})$ .

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Step 3.9.2.1.1.3.1

Multiply $- 1$ by $2$ .

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} - 2 (\frac{71}{125})$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.3.2

Combine $- 2$ and $\frac{71}{125}$ .

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} + \frac{- 2 \cdot 71}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.3.3

Multiply $- 2$ by $71$ .

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} + \frac{- 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} + \frac{- 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.1.4

Move the negative in front of the fraction.

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = 3 B + \frac{6}{5} + \frac{4 B}{25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.2

To write $3 B$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$0 = 3 B \cdot \frac{25}{25} + \frac{4 B}{25} + \frac{6}{5} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.3

Combine $3 B$ and $\frac{25}{25}$ .

$0 = \frac{3 B \cdot 25}{25} + \frac{4 B}{25} + \frac{6}{5} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.4

Combine the numerators over the common denominator.

$0 = \frac{3 B \cdot 25 + 4 B}{25} + \frac{6}{5} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5

Find the common denominator.

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Step 3.9.2.1.5.1

Multiply $\frac{3 B \cdot 25 + 4 B}{25}$ by $\frac{5}{5}$ .

$0 = \frac{3 B \cdot 25 + 4 B}{25} \cdot \frac{5}{5} + \frac{6}{5} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.2

Multiply $\frac{3 B \cdot 25 + 4 B}{25}$ by $\frac{5}{5}$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{25 \cdot 5} + \frac{6}{5} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.3

Multiply $\frac{6}{5}$ by $\frac{25}{25}$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{25 \cdot 5} + \frac{6}{5} \cdot \frac{25}{25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.4

Multiply $\frac{6}{5}$ by $\frac{25}{25}$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{25 \cdot 5} + \frac{6 \cdot 25}{5 \cdot 25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.5

Reorder the factors of $25 \cdot 5$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{5 \cdot 25} + \frac{6 \cdot 25}{5 \cdot 25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.6

Multiply $5$ by $25$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{125} + \frac{6 \cdot 25}{5 \cdot 25} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.5.7

Multiply $5$ by $25$ .

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{125} + \frac{6 \cdot 25}{125} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5}{125} + \frac{6 \cdot 25}{125} - \frac{142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.6

Combine the numerators over the common denominator.

$0 = \frac{(3 B \cdot 25 + 4 B) \cdot 5 + 6 \cdot 25 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.7

Simplify each term.

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Step 3.9.2.1.7.1

Multiply $25$ by $3$ .

$0 = \frac{(75 B + 4 B) \cdot 5 + 6 \cdot 25 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.7.2

Add $75 B$ and $4 B$ .

$0 = \frac{79 B \cdot 5 + 6 \cdot 25 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.7.3

Multiply $5$ by $79$ .

$0 = \frac{395 B + 6 \cdot 25 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.7.4

Multiply $6$ by $25$ .

$0 = \frac{395 B + 150 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = \frac{395 B + 150 - 142}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.9.2.1.8

Subtract $142$ from $150$ .

$0 = \frac{395 B + 8}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = \frac{395 B + 8}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = \frac{395 B + 8}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$0 = \frac{395 B + 8}{125}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.10

Solve for $B$ in $0 = \frac{395 B + 8}{125}$ .

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Step 3.10.1

Set the numerator equal to zero.

$395 B + 8 = 0$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.10.2

Solve the equation for $B$ .

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Step 3.10.2.1

Subtract $8$ from both sides of the equation.

$395 B = - 8$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.10.2.2

Divide each term in $395 B = - 8$ by $395$ and simplify.

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Step 3.10.2.2.1

Divide each term in $395 B = - 8$ by $395$ .

$\frac{395 B}{395} = \frac{- 8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

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 $395$ .

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Step 3.10.2.2.2.1.1

Cancel the common factor.

$\frac{395 B}{395} = \frac{- 8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.10.2.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = \frac{- 8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = \frac{- 8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

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.

$B = - \frac{8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = - \frac{8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = - \frac{8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = - \frac{8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$B = - \frac{8}{395}$

$C = \frac{2 B}{25} - \frac{71}{125}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11

Replace all occurrences of $B$ with $- \frac{8}{395}$ in each equation.

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Step 3.11.1

Replace all occurrences of $B$ in $C = \frac{2 B}{25} - \frac{71}{125}$ with $- \frac{8}{395}$ .

$C = \frac{2 (- \frac{8}{395})}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2

Simplify the right side.

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Step 3.11.2.1

Simplify $\frac{2 (- \frac{8}{395})}{25} - \frac{71}{125}$ .

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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 $2$ .

$C = \frac{- 2 (\frac{8}{395})}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.1.2

Combine $- 2$ and $\frac{8}{395}$ .

$C = \frac{\frac{- 2 \cdot 8}{395}}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{\frac{- 2 \cdot 8}{395}}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.2

Multiply $- 2$ by $8$ .

$C = \frac{\frac{- 16}{395}}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.3

Move the negative in front of the fraction.

$C = \frac{- \frac{16}{395}}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.4

Multiply the numerator by the reciprocal of the denominator.

$C = - \frac{16}{395} \cdot \frac{1}{25} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.5

Multiply $- \frac{16}{395} \cdot \frac{1}{25}$ .

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Step 3.11.2.1.1.5.1

Multiply $\frac{1}{25}$ by $\frac{16}{395}$ .

$C = - \frac{16}{25 \cdot 395} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.1.5.2

Multiply $25$ by $395$ .

$C = - \frac{16}{9875} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = - \frac{16}{9875} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = - \frac{16}{9875} - \frac{71}{125}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.2

To write $- \frac{71}{125}$ as a fraction with a common denominator, multiply by $\frac{79}{79}$ .

$C = - \frac{16}{9875} - \frac{71}{125} \cdot \frac{79}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.3

Write each expression with a common denominator of $9875$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.11.2.1.3.1

Multiply $\frac{71}{125}$ by $\frac{79}{79}$ .

$C = - \frac{16}{9875} - \frac{71 \cdot 79}{125 \cdot 79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.3.2

Multiply $125$ by $79$ .

$C = - \frac{16}{9875} - \frac{71 \cdot 79}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = - \frac{16}{9875} - \frac{71 \cdot 79}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.4

Combine the numerators over the common denominator.

$C = \frac{- 16 - 71 \cdot 79}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.5

Simplify the numerator.

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Step 3.11.2.1.5.1

Multiply $- 71$ by $79$ .

$C = \frac{- 16 - 5609}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.5.2

Subtract $5609$ from $- 16$ .

$C = \frac{- 5625}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{- 5625}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.6

Cancel the common factor of $- 5625$ and $9875$ .

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Step 3.11.2.1.6.1

Factor $125$ out of $- 5625$ .

$C = \frac{125 (- 45)}{9875}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.6.2

Cancel the common factors.

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Step 3.11.2.1.6.2.1

Factor $125$ out of $9875$ .

$C = \frac{125 \cdot - 45}{125 \cdot 79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.6.2.2

Cancel the common factor.

$C = \frac{125 \cdot - 45}{125 \cdot 79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.6.2.3

Rewrite the expression.

$C = \frac{- 45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{- 45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = \frac{- 45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.2.1.7

Move the negative in front of the fraction.

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

$D = - \frac{B}{5} - \frac{2}{25}$

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

Step 3.11.3

Replace all occurrences of $B$ in $D = - \frac{B}{5} - \frac{2}{25}$ with $- \frac{8}{395}$ .

$D = - \frac{- \frac{8}{395}}{5} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4

Simplify the right side.

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Step 3.11.4.1

Simplify $- \frac{- \frac{8}{395}}{5} - \frac{2}{25}$ .

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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.

$D = - (- \frac{8}{395} \cdot \frac{1}{5}) - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.1.2

Multiply $- \frac{8}{395} \cdot \frac{1}{5}$ .

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Step 3.11.4.1.1.2.1

Multiply $\frac{1}{5}$ by $\frac{8}{395}$ .

$D = \frac{8}{5 \cdot 395} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.1.2.2

Multiply $5$ by $395$ .

$D = \frac{8}{1975} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{8}{1975} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.1.3

Multiply $- - \frac{8}{1975}$ .

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Step 3.11.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$D = 1 (\frac{8}{1975}) - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.1.3.2

Multiply $\frac{8}{1975}$ by $1$ .

$D = \frac{8}{1975} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{8}{1975} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{8}{1975} - \frac{2}{25}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.2

To write $- \frac{2}{25}$ as a fraction with a common denominator, multiply by $\frac{79}{79}$ .

$D = \frac{8}{1975} - \frac{2}{25} \cdot \frac{79}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.3

Write each expression with a common denominator of $1975$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.11.4.1.3.1

Multiply $\frac{2}{25}$ by $\frac{79}{79}$ .

$D = \frac{8}{1975} - \frac{2 \cdot 79}{25 \cdot 79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.3.2

Multiply $25$ by $79$ .

$D = \frac{8}{1975} - \frac{2 \cdot 79}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{8}{1975} - \frac{2 \cdot 79}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.4

Combine the numerators over the common denominator.

$D = \frac{8 - 2 \cdot 79}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.5

Simplify the numerator.

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Step 3.11.4.1.5.1

Multiply $- 2$ by $79$ .

$D = \frac{8 - 158}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.5.2

Subtract $158$ from $8$ .

$D = \frac{- 150}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{- 150}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.6

Cancel the common factor of $- 150$ and $1975$ .

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Step 3.11.4.1.6.1

Factor $25$ out of $- 150$ .

$D = \frac{25 (- 6)}{1975}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.6.2

Cancel the common factors.

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Step 3.11.4.1.6.2.1

Factor $25$ out of $1975$ .

$D = \frac{25 \cdot - 6}{25 \cdot 79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.6.2.2

Cancel the common factor.

$D = \frac{25 \cdot - 6}{25 \cdot 79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.6.2.3

Rewrite the expression.

$D = \frac{- 6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{- 6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = \frac{- 6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.11.4.1.7

Move the negative in front of the fraction.

$D = - \frac{6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = - \frac{6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = - \frac{6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

$D = - \frac{6}{79}$

$C = - \frac{45}{79}$

$B = - \frac{8}{395}$

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

Step 3.12

List all of the solutions.

$D = - \frac{6}{79}, C = - \frac{45}{79}, B = - \frac{8}{395}, A = \frac{1}{5}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{2 x + 5} + \frac{C x + D}{3 x^{2} + 1}$ with the values found for $A$ , $B$ , $C$ , and $D$ .

$\frac{\frac{1}{5}}{x} + \frac{- \frac{8}{395}}{2 x + 5} + \frac{- \frac{45}{79 x} - \frac{6}{79}}{3 x^{2} + 1}$

Step 5

Simplify each term.

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Step 5.1

Combine $x$ and $\frac{45}{79}$ .

$\frac{\frac{1}{5}}{x} + \frac{- \frac{8}{395}}{2 x + 5} + \frac{- \frac{x \cdot 45}{79} - \frac{6}{79}}{3 x^{2} + 1}$

Step 5.2

Move $45$ to the left of $x$ .

$\frac{\frac{1}{5}}{x} + \frac{- \frac{8}{395}}{2 x + 5} + \frac{- \frac{45 x}{79} - \frac{6}{79}}{3 x^{2} + 1}$