Solve by Factoring 1/(2x-1)-1/(2x+1)=1/12

$\frac{1}{2 x - 1} - \frac{1}{2 x + 1} = \frac{1}{12}$

Step 1

Subtract $\frac{1}{12}$ from both sides of the equation.

$\frac{1}{2 x - 1} - \frac{1}{2 x + 1} - \frac{1}{12} = 0$

Step 2

Simplify $\frac{1}{2 x - 1} - \frac{1}{2 x + 1} - \frac{1}{12}$ .

To write $\frac{1}{2 x - 1}$ as a fraction with a common denominator, multiply by $\frac{2 x + 1}{2 x + 1}$ .

$\frac{1}{2 x - 1} \cdot \frac{2 x + 1}{2 x + 1} - \frac{1}{2 x + 1} - \frac{1}{12} = 0$

To write $- \frac{1}{2 x + 1}$ as a fraction with a common denominator, multiply by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{1}{2 x - 1} \cdot \frac{2 x + 1}{2 x + 1} - \frac{1}{2 x + 1} \cdot \frac{2 x - 1}{2 x - 1} - \frac{1}{12} = 0$

Write each expression with a common denominator of $(2 x - 1) (2 x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{2 x - 1}$ by $\frac{2 x + 1}{2 x + 1}$ .

$\frac{2 x + 1}{(2 x - 1) (2 x + 1)} - \frac{1}{2 x + 1} \cdot \frac{2 x - 1}{2 x - 1} - \frac{1}{12} = 0$

Multiply $\frac{1}{2 x + 1}$ by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{2 x + 1}{(2 x - 1) (2 x + 1)} - \frac{2 x - 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Reorder the factors of $(2 x - 1) (2 x + 1)$ .

$\frac{2 x + 1}{(2 x + 1) (2 x - 1)} - \frac{2 x - 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Combine the numerators over the common denominator.

$\frac{2 x + 1 - (2 x - 1)}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{2 x + 1 - (2 x) - - 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Multiply $2$ by $- 1$ .

$\frac{2 x + 1 - 2 x - - 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Multiply $- 1$ by $- 1$ .

$\frac{2 x + 1 - 2 x + 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Subtract $2 x$ from $2 x$ .

$\frac{0 + 1 + 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Add $0$ and $1$ .

$\frac{1 + 1}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Add $1$ and $1$ .

$\frac{2}{(2 x + 1) (2 x - 1)} - \frac{1}{12} = 0$

Step 3

To write $\frac{2}{(2 x + 1) (2 x - 1)}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$\frac{2}{(2 x + 1) (2 x - 1)} \cdot \frac{12}{12} - \frac{1}{12} = 0$

Step 4

To write $- \frac{1}{12}$ as a fraction with a common denominator, multiply by $\frac{(2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1)}$ .

$\frac{2}{(2 x + 1) (2 x - 1)} \cdot \frac{12}{12} - \frac{1}{12} \cdot \frac{(2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1)} = 0$

Step 5

Write each expression with a common denominator of $(2 x + 1) (2 x - 1) \cdot 12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{(2 x + 1) (2 x - 1)}$ by $\frac{12}{12}$ .

$\frac{2 \cdot 12}{(2 x + 1) (2 x - 1) \cdot 12} - \frac{1}{12} \cdot \frac{(2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1)} = 0$

Multiply $\frac{1}{12}$ by $\frac{(2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1)}$ .

$\frac{2 \cdot 12}{(2 x + 1) (2 x - 1) \cdot 12} - \frac{(2 x + 1) (2 x - 1)}{12 ((2 x + 1) (2 x - 1))} = 0$

Reorder the factors of $(2 x + 1) (2 x - 1) \cdot 12$ .

$\frac{2 \cdot 12}{12 (2 x + 1) (2 x - 1)} - \frac{(2 x + 1) (2 x - 1)}{12 ((2 x + 1) (2 x - 1))} = 0$

Reorder the factors of $12 ((2 x + 1) (2 x - 1))$ .

$\frac{2 \cdot 12}{12 (2 x + 1) (2 x - 1)} - \frac{(2 x + 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Step 6

Combine the numerators over the common denominator.

$\frac{2 \cdot 12 - (2 x + 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Step 7

Simplify the numerator.

Multiply $2$ by $12$ .

$\frac{24 - (2 x + 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Apply the distributive property.

$\frac{24 + (- (2 x) - 1 \cdot 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $2$ by $- 1$ .

$\frac{24 + (- 2 x - 1 \cdot 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $- 1$ by $1$ .

$\frac{24 + (- 2 x - 1) (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

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

Apply the distributive property.

$\frac{24 - 2 x (2 x - 1) - 1 (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Apply the distributive property.

$\frac{24 - 2 x (2 x) - 2 x \cdot - 1 - 1 (2 x - 1)}{12 (2 x + 1) (2 x - 1)} = 0$

Apply the distributive property.

$\frac{24 - 2 x (2 x) - 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$\frac{24 - 2 \cdot (2 x \cdot x) - 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

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

Move $x$ .

$\frac{24 - 2 \cdot (2 (x \cdot x)) - 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $x$ by $x$ .

$\frac{24 - 2 \cdot (2 x^{2}) - 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $- 2$ by $2$ .

$\frac{24 - 4 x^{2} - 2 x \cdot - 1 - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $- 1$ by $- 2$ .

$\frac{24 - 4 x^{2} + 2 x - 1 (2 x) - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $2$ by $- 1$ .

$\frac{24 - 4 x^{2} + 2 x - 2 x - 1 \cdot - 1}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $- 1$ by $- 1$ .

$\frac{24 - 4 x^{2} + 2 x - 2 x + 1}{12 (2 x + 1) (2 x - 1)} = 0$

Subtract $2 x$ from $2 x$ .

$\frac{24 - 4 x^{2} + 0 + 1}{12 (2 x + 1) (2 x - 1)} = 0$

Add $- 4 x^{2}$ and $0$ .

$\frac{24 - 4 x^{2} + 1}{12 (2 x + 1) (2 x - 1)} = 0$

Add $24$ and $1$ .

$\frac{- 4 x^{2} + 25}{12 (2 x + 1) (2 x - 1)} = 0$

Rewrite $- 4 x^{2} + 25$ in a factored form.

Rewrite $25$ as $5^{2}$ .

$\frac{- 4 x^{2} + 5^{2}}{12 (2 x + 1) (2 x - 1)} = 0$

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{- {(2 x)}^{2} + 5^{2}}{12 (2 x + 1) (2 x - 1)} = 0$

Reorder $- {(2 x)}^{2}$ and $5^{2}$ .

$\frac{5^{2} - {(2 x)}^{2}}{12 (2 x + 1) (2 x - 1)} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = 2 x$ .

$\frac{(5 + 2 x) (5 - (2 x))}{12 (2 x + 1) (2 x - 1)} = 0$

Multiply $2$ by $- 1$ .

$\frac{(5 + 2 x) (5 - 2 x)}{12 (2 x + 1) (2 x - 1)} = 0$

Step 8

Set the numerator equal to zero.

$(5 + 2 x) (5 - 2 x) = 0$

Step 9

Solve the equation for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 + 2 x = 0$

$5 - 2 x = 0$

Set $5 + 2 x$ equal to $0$ and solve for $x$ .

Set $5 + 2 x$ equal to $0$ .

$5 + 2 x = 0$

Solve $5 + 2 x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Set $5 - 2 x$ equal to $0$ and solve for $x$ .

Set $5 - 2 x$ equal to $0$ .

$5 - 2 x = 0$

Solve $5 - 2 x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- 2 x = - 5$

Divide each term in $- 2 x = - 5$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 5$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 5}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 5}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{5}{2}$

The final solution is all the values that make $(5 + 2 x) (5 - 2 x) = 0$ true.

$x = - \frac{5}{2}, \frac{5}{2}$

Solve by Factoring 1/(2x-1)-1/(2x+1)=1/12

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