Basic Math Examples

$\frac{6 b - 12}{8 b^{2} + 29 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1

Factor each term.

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Step 1.1

Factor $6$ out of $6 b - 12$ .

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Step 1.1.1

Factor $6$ out of $6 b$ .

$\frac{6 (b) - 12}{8 b^{2} + 29 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.1.2

Factor $6$ out of $- 12$ .

$\frac{6 b + 6 \cdot - 2}{8 b^{2} + 29 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.1.3

Factor $6$ out of $6 b + 6 \cdot - 2$ .

$\frac{6 (b - 2)}{8 b^{2} + 29 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2

Factor by grouping.

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Step 1.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 8 \cdot 15 = 120$ and whose sum is $b = 29$ .

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Step 1.2.1.1

Factor $29$ out of $29 b$ .

$\frac{6 (b - 2)}{8 b^{2} + 29 (b) + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2.1.2

Rewrite $29$ as $5$ plus $24$

$\frac{6 (b - 2)}{8 b^{2} + (5 + 24) b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2.1.3

Apply the distributive property.

$\frac{6 (b - 2)}{8 b^{2} + 5 b + 24 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$\frac{6 (b - 2)}{(8 b^{2} + 5 b) + 24 b + 15} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{6 (b - 2)}{b (8 b + 5) + 3 (8 b + 5)} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $8 b + 5$ .

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 b + 15}$

Step 1.3

Factor by grouping.

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Step 1.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 8 \cdot 15 = 120$ and whose sum is $b = 29$ .

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Step 1.3.1.1

Factor $29$ out of $29 b$ .

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 29 (b) + 15}$

Step 1.3.1.2

Rewrite $29$ as $5$ plus $24$

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + (5 + 24) b + 15}$

Step 1.3.1.3

Apply the distributive property.

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{8 b^{2} + 5 b + 24 b + 15}$

Step 1.3.2

Factor out the greatest common factor from each group.

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Step 1.3.2.1

Group the first two terms and the last two terms.

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{(8 b^{2} + 5 b) + 24 b + 15}$

Step 1.3.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{b (8 b + 5) + 3 (8 b + 5)}$

Step 1.3.3

Factor the polynomial by factoring out the greatest common factor, $8 b + 5$ .

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{(8 b + 5) (b + 3)}$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(8 b + 5) (b + 3), 8 b + 5, (8 b + 5) (b + 3)$

Step 2.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.5

The factor for $8 b + 5$ is $8 b + 5$ itself.

$(8 b + 5) = 8 b + 5$

$(8 b + 5)$ occurs $1$ time.

Step 2.6

The factor for $b + 3$ is $b + 3$ itself.

$(b + 3) = b + 3$

$(b + 3)$ occurs $1$ time.

Step 2.7

The factor for $8 b + 5$ is $8 b + 5$ itself.

$(8 b + 5) = 8 b + 5$

$(8 b + 5)$ occurs $1$ time.

Step 2.8

The factor for $b + 3$ is $b + 3$ itself.

$(b + 3) = b + 3$

$(b + 3)$ occurs $1$ time.

Step 2.9

The LCM of $8 b + 5, b + 3, 8 b + 5, 8 b + 5, b + 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(8 b + 5) (b + 3)$

Step 3

Multiply each term in $\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{(8 b + 5) (b + 3)}$ by $(8 b + 5) (b + 3)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{6 (b - 2)}{(8 b + 5) (b + 3)} = \frac{1}{8 b + 5} + \frac{1}{(8 b + 5) (b + 3)}$ by $(8 b + 5) (b + 3)$ .

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3)) = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $(8 b + 5) (b + 3)$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{6 (b - 2)}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3)) = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.2.1.2

Rewrite the expression.

$6 (b - 2) = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.2.2

Apply the distributive property.

$6 b + 6 \cdot - 2 = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.2.3

Multiply $6$ by $- 2$ .

$6 b - 12 = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Cancel the common factor of $8 b + 5$ .

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Step 3.3.1.1.1

Cancel the common factor.

$6 b - 12 = \frac{1}{8 b + 5} ((8 b + 5) (b + 3)) + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.3.1.1.2

Rewrite the expression.

$6 b - 12 = b + 3 + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.3.1.2

Cancel the common factor of $(8 b + 5) (b + 3)$ .

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Step 3.3.1.2.1

Cancel the common factor.

$6 b - 12 = b + 3 + \frac{1}{(8 b + 5) (b + 3)} ((8 b + 5) (b + 3))$

Step 3.3.1.2.2

Rewrite the expression.

$6 b - 12 = b + 3 + 1$

Step 3.3.2

Add $3$ and $1$ .

$6 b - 12 = b + 4$

Step 4

Solve the equation.

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Step 4.1

Move all terms containing $b$ to the left side of the equation.

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Step 4.1.1

Subtract $b$ from both sides of the equation.

$6 b - 12 - b = 4$

Step 4.1.2

Subtract $b$ from $6 b$ .

$5 b - 12 = 4$

Step 4.2

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

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Step 4.2.1

Add $12$ to both sides of the equation.

$5 b = 4 + 12$

Step 4.2.2

Add $4$ and $12$ .

$5 b = 16$

Step 4.3

Divide each term in $5 b = 16$ by $5$ and simplify.

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Step 4.3.1

Divide each term in $5 b = 16$ by $5$ .

$\frac{5 b}{5} = \frac{16}{5}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $5$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{5 b}{5} = \frac{16}{5}$

Step 4.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{16}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$b = \frac{16}{5}$

Decimal Form:

$b = 3.2$

Mixed Number Form:

$b = 3 \frac{1}{5}$