Solve for x 2+3/((2x+1)(2x-1))=3

$2 + \frac{3}{(2 x + 1) (2 x - 1)} = 3$

Find the LCD of the terms in the equation.

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

$1, (2 x + 1) (2 x - 1), 1$

The LCM of one and any expression is the expression.

$(2 x + 1) (2 x - 1)$

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

Multiply each term in $2 + \frac{3}{(2 x + 1) (2 x - 1)} = 3$ by $(2 x + 1) (2 x - 1)$ .

$2 ((2 x + 1) (2 x - 1)) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Simplify the left side.

Simplify each term.

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

Apply the distributive property.

$2 (2 x (2 x - 1) + 1 (2 x - 1)) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Apply the distributive property.

$2 (2 x (2 x) + 2 x \cdot - 1 + 1 (2 x - 1)) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Apply the distributive property.

$2 (2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

$2 (2 x (2 x) - 1 \cdot 2 x + 1 \cdot 2 x + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$2 (2 x (2 x) + 0 + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Add $2 x (2 x)$ and $0$ .

$2 (2 x (2 x) + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 (2 \cdot 2 x \cdot x + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

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

Move $x$ .

$2 (2 \cdot 2 (x \cdot x) + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Multiply $x$ by $x$ .

$2 (2 \cdot 2 x^{2} + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Multiply $2$ by $2$ .

$2 (4 x^{2} + 1 \cdot - 1) + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Multiply $- 1$ by $1$ .

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

Apply the distributive property.

$2 (4 x^{2}) + 2 \cdot - 1 + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Multiply $4$ by $2$ .

$8 x^{2} + 2 \cdot - 1 + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Multiply $2$ by $- 1$ .

$8 x^{2} - 2 + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Cancel the common factor of $(2 x + 1) (2 x - 1)$ .

Cancel the common factor.

$8 x^{2} - 2 + \frac{3}{(2 x + 1) (2 x - 1)} ((2 x + 1) (2 x - 1)) = 3 ((2 x + 1) (2 x - 1))$

Rewrite the expression.

$8 x^{2} - 2 + 3 = 3 ((2 x + 1) (2 x - 1))$

Add $- 2$ and $3$ .

$8 x^{2} + 1 = 3 ((2 x + 1) (2 x - 1))$

Simplify the right side.

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

Apply the distributive property.

$8 x^{2} + 1 = 3 (2 x (2 x - 1) + 1 (2 x - 1))$

Apply the distributive property.

$8 x^{2} + 1 = 3 (2 x (2 x) + 2 x \cdot - 1 + 1 (2 x - 1))$

Apply the distributive property.

$8 x^{2} + 1 = 3 (2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1)$

Simplify terms.

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

$8 x^{2} + 1 = 3 (2 x (2 x) - 1 \cdot 2 x + 1 \cdot 2 x + 1 \cdot - 1)$

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$8 x^{2} + 1 = 3 (2 x (2 x) + 0 + 1 \cdot - 1)$

Add $2 x (2 x)$ and $0$ .

$8 x^{2} + 1 = 3 (2 x (2 x) + 1 \cdot - 1)$

Simplify each term.

Rewrite using the commutative property of multiplication.

$8 x^{2} + 1 = 3 (2 \cdot 2 x \cdot x + 1 \cdot - 1)$

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

Move $x$ .

$8 x^{2} + 1 = 3 (2 \cdot 2 (x \cdot x) + 1 \cdot - 1)$

Multiply $x$ by $x$ .

$8 x^{2} + 1 = 3 (2 \cdot 2 x^{2} + 1 \cdot - 1)$

Multiply $2$ by $2$ .

$8 x^{2} + 1 = 3 (4 x^{2} + 1 \cdot - 1)$

Multiply $- 1$ by $1$ .

$8 x^{2} + 1 = 3 (4 x^{2} - 1)$

Simplify by multiplying through.

Apply the distributive property.

$8 x^{2} + 1 = 3 (4 x^{2}) + 3 \cdot - 1$

Multiply.

Multiply $4$ by $3$ .

$8 x^{2} + 1 = 12 x^{2} + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$8 x^{2} + 1 = 12 x^{2} - 3$

Solve the equation.

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

Subtract $12 x^{2}$ from both sides of the equation.

$8 x^{2} + 1 - 12 x^{2} = - 3$

Subtract $12 x^{2}$ from $8 x^{2}$ .

$- 4 x^{2} + 1 = - 3$

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

Subtract $1$ from both sides of the equation.

$- 4 x^{2} = - 3 - 1$

Subtract $1$ from $- 3$ .

$- 4 x^{2} = - 4$

Divide each term in $- 4 x^{2} = - 4$ by $- 4$ and simplify.

Divide each term in $- 4 x^{2} = - 4$ by $- 4$ .

$\frac{- 4 x^{2}}{- 4} = \frac{- 4}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x^{2}}{- 4} = \frac{- 4}{- 4}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{- 4}$

Simplify the right side.

Divide $- 4$ by $- 4$ .

$x^{2} = 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Solve for x 2+3/((2x+1)(2x-1))=3

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