Solve for x 1/(2x-1)-1/(2x+1)=1/40

$\frac{1}{2 x - 1} - \frac{1}{2 x + 1} = \frac{1}{40}$

Step 1

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.

$2 x - 1, 2 x + 1, 40$

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.

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

Not prime

The prime factors for $40$ are $2 \cdot 2 \cdot 2 \cdot 5$ .

$40$ has factors of $2$ and $20$ .

$2 \cdot 20$

$20$ has factors of $2$ and $10$ .

$2 \cdot 2 \cdot 10$

$10$ has factors of $2$ and $5$ .

$2 \cdot 2 \cdot 2 \cdot 5$

Multiply $2 \cdot 2 \cdot 2 \cdot 5$ .

Multiply $2$ by $2$ .

$4 \cdot 2 \cdot 5$

Multiply $4$ by $2$ .

$8 \cdot 5$

Multiply $8$ by $5$ .

$40$

The factor for $2 x - 1$ is $2 x - 1$ itself.

$(2 x - 1) = 2 x - 1$

$(2 x - 1)$ occurs $1$ time.

The factor for $2 x + 1$ is $2 x + 1$ itself.

$(2 x + 1) = 2 x + 1$

$(2 x + 1)$ occurs $1$ time.

The LCM of $2 x - 1, 2 x + 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

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

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$40 (2 x - 1) (2 x + 1)$

Step 2

Multiply each term in $\frac{1}{2 x - 1} - \frac{1}{2 x + 1} = \frac{1}{40}$ by $40 (2 x - 1) (2 x + 1)$ to eliminate the fractions.

Multiply each term in $\frac{1}{2 x - 1} - \frac{1}{2 x + 1} = \frac{1}{40}$ by $40 (2 x - 1) (2 x + 1)$ .

$\frac{1}{2 x - 1} (40 (2 x - 1) (2 x + 1)) - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$40 \frac{1}{2 x - 1} ((2 x - 1) (2 x + 1)) - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Combine $40$ and $\frac{1}{2 x - 1}$ .

$\frac{40}{2 x - 1} ((2 x - 1) (2 x + 1)) - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Cancel the common factor of $2 x - 1$ .

Cancel the common factor.

$\frac{40}{2 x - 1} ((2 x - 1) (2 x + 1)) - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Rewrite the expression.

$40 (2 x + 1) - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Apply the distributive property.

$40 (2 x) + 40 \cdot 1 - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Multiply $2$ by $40$ .

$80 x + 40 \cdot 1 - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Multiply $40$ by $1$ .

$80 x + 40 - \frac{1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Cancel the common factor of $2 x + 1$ .

Move the leading negative in $- \frac{1}{2 x + 1}$ into the numerator.

$80 x + 40 + \frac{- 1}{2 x + 1} (40 (2 x - 1) (2 x + 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Factor $2 x + 1$ out of $40 (2 x - 1) (2 x + 1)$ .

$80 x + 40 + \frac{- 1}{2 x + 1} ((2 x + 1) (40 (2 x - 1))) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Cancel the common factor.

$80 x + 40 + \frac{- 1}{2 x + 1} ((2 x + 1) (40 (2 x - 1))) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Rewrite the expression.

$80 x + 40 - (40 (2 x - 1)) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Multiply $40$ by $- 1$ .

$80 x + 40 - 40 (2 x - 1) = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Apply the distributive property.

$80 x + 40 - 40 (2 x) - 40 \cdot - 1 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Multiply $2$ by $- 40$ .

$80 x + 40 - 80 x - 40 \cdot - 1 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Multiply $- 40$ by $- 1$ .

$80 x + 40 - 80 x + 40 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Simplify by adding terms.

Combine the opposite terms in $80 x + 40 - 80 x + 40$ .

Subtract $80 x$ from $80 x$ .

$0 + 40 + 40 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Add $0$ and $40$ .

$40 + 40 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Add $40$ and $40$ .

$80 = \frac{1}{40} (40 (2 x - 1) (2 x + 1))$

Simplify the right side.

Cancel the common factor of $40$ .

Factor $40$ out of $40 (2 x - 1) (2 x + 1)$ .

$80 = \frac{1}{40} (40 ((2 x - 1) (2 x + 1)))$

Cancel the common factor.

$80 = \frac{1}{40} (40 ((2 x - 1) (2 x + 1)))$

Rewrite the expression.

$80 = (2 x - 1) (2 x + 1)$

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

Apply the distributive property.

$80 = 2 x (2 x + 1) - 1 (2 x + 1)$

Apply the distributive property.

$80 = 2 x (2 x) + 2 x \cdot 1 - 1 (2 x + 1)$

Apply the distributive property.

$80 = 2 x (2 x) + 2 x \cdot 1 - 1 (2 x) - 1 \cdot 1$

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 $- 1 (2 x)$ .

$80 = 2 x (2 x) + 1 \cdot 2 x - 1 \cdot 2 x - 1 \cdot 1$

Subtract $2 x$ from $1 \cdot 2 x$ .

$80 = 2 x (2 x) + 0 - 1 \cdot 1$

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

$80 = 2 x (2 x) - 1 \cdot 1$

Simplify each term.

Rewrite using the commutative property of multiplication.

$80 = 2 \cdot 2 x \cdot x - 1 \cdot 1$

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

Move $x$ .

$80 = 2 \cdot 2 (x \cdot x) - 1 \cdot 1$

Multiply $x$ by $x$ .

$80 = 2 \cdot 2 x^{2} - 1 \cdot 1$

Multiply $2$ by $2$ .

$80 = 4 x^{2} - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$80 = 4 x^{2} - 1$

Step 3

Solve the equation.

Rewrite the equation as $4 x^{2} - 1 = 80$ .

$4 x^{2} - 1 = 80$

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

Add $1$ to both sides of the equation.

$4 x^{2} = 80 + 1$

Add $80$ and $1$ .

$4 x^{2} = 81$

Divide each term in $4 x^{2} = 81$ by $4$ and simplify.

Divide each term in $4 x^{2} = 81$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{81}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{81}{4}$

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

$x^{2} = \frac{81}{4}$

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

$x = \pm \sqrt{\frac{81}{4}}$

Simplify $\pm \sqrt{\frac{81}{4}}$ .

Rewrite $\sqrt{\frac{81}{4}}$ as $\frac{\sqrt{81}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{81}}{\sqrt{4}}$

Simplify the numerator.

Rewrite $81$ as $9^{2}$ .

$x = \pm \frac{\sqrt{9^{2}}}{\sqrt{4}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{9}{\sqrt{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$x = \pm \frac{9}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{9}{2}$

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.

$x = \frac{9}{2}$

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

$x = - \frac{9}{2}$

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

$x = \frac{9}{2}, - \frac{9}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{9}{2}, - \frac{9}{2}$

Decimal Form:

$x = 4.5, - 4.5$

Mixed Number Form:

$x = 4 \frac{1}{2}, - 4 \frac{1}{2}$

Solve for x 1/(2x-1)-1/(2x+1)=1/40

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