Solve for x 1/(3x-3)+1/(4x+4)=1/(12x-12)

$\frac{1}{3 x - 3} + \frac{1}{4 x + 4} = \frac{1}{12 x - 12}$

Step 1

Factor each term.

Factor $3$ out of $3 x - 3$ .

Factor $3$ out of $3 x$ .

$\frac{1}{3 (x) - 3} + \frac{1}{4 x + 4} = \frac{1}{12 x - 12}$

Factor $3$ out of $- 3$ .

$\frac{1}{3 (x) + 3 (- 1)} + \frac{1}{4 x + 4} = \frac{1}{12 x - 12}$

Factor $3$ out of $3 (x) + 3 (- 1)$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 x + 4} = \frac{1}{12 x - 12}$

Factor $4$ out of $4 x + 4$ .

Factor $4$ out of $4 x$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x) + 4} = \frac{1}{12 x - 12}$

Factor $4$ out of $4$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x) + 4 (1)} = \frac{1}{12 x - 12}$

Factor $4$ out of $4 (x) + 4 (1)$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 x - 12}$

Factor $12$ out of $12 x - 12$ .

Factor $12$ out of $12 x$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 (x) - 12}$

Factor $12$ out of $- 12$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 (x) + 12 (- 1)}$

Factor $12$ out of $12 (x) + 12 (- 1)$ .

$\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 (x - 1)}$

Step 2

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.

$3 (x - 1), 4 (x + 1), 12 (x - 1)$

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.

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

The prime factors for $12$ are $2 \cdot 2 \cdot 3$ .

$12$ has factors of $2$ and $6$ .

$2 \cdot 6$

$6$ has factors of $2$ and $3$ .

$2 \cdot 2 \cdot 3$

Multiply $2 \cdot 2 \cdot 3$ .

Multiply $2$ by $2$ .

$4 \cdot 3$

Multiply $4$ by $3$ .

$12$

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

$(x - 1) = x - 1$

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

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

$(x + 1) = x + 1$

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

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

$(x - 1) = x - 1$

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

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

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

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

$12 (x - 1) (x + 1)$

Step 3

Multiply each term in $\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 (x - 1)}$ by $12 (x - 1) (x + 1)$ to eliminate the fractions.

Multiply each term in $\frac{1}{3 (x - 1)} + \frac{1}{4 (x + 1)} = \frac{1}{12 (x - 1)}$ by $12 (x - 1) (x + 1)$ .

$\frac{1}{3 (x - 1)} (12 (x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$12 \frac{1}{3 (x - 1)} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor of $3$ .

Factor $3$ out of $12$ .

$3 (4) \frac{1}{3 (x - 1)} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor.

$3 \cdot 4 \frac{1}{3 (x - 1)} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Rewrite the expression.

$4 \frac{1}{x - 1} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Combine $4$ and $\frac{1}{x - 1}$ .

$\frac{4}{x - 1} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{4}{x - 1} ((x - 1) (x + 1)) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Rewrite the expression.

$4 (x + 1) + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Apply the distributive property.

$4 x + 4 \cdot 1 + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Multiply $4$ by $1$ .

$4 x + 4 + \frac{1}{4 (x + 1)} (12 (x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Rewrite using the commutative property of multiplication.

$4 x + 4 + 12 \frac{1}{4 (x + 1)} ((x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

$4 x + 4 + 4 (3) \frac{1}{4 (x + 1)} ((x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor.

$4 x + 4 + 4 \cdot 3 \frac{1}{4 (x + 1)} ((x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Rewrite the expression.

$4 x + 4 + 3 \frac{1}{x + 1} ((x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Combine $3$ and $\frac{1}{x + 1}$ .

$4 x + 4 + \frac{3}{x + 1} ((x - 1) (x + 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor of $x + 1$ .

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

$4 x + 4 + \frac{3}{x + 1} ((x + 1) (x - 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Cancel the common factor.

$4 x + 4 + \frac{3}{x + 1} ((x + 1) (x - 1)) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Rewrite the expression.

$4 x + 4 + 3 (x - 1) = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Apply the distributive property.

$4 x + 4 + 3 x + 3 \cdot - 1 = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Multiply $3$ by $- 1$ .

$4 x + 4 + 3 x - 3 = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Simplify by adding terms.

Add $4 x$ and $3 x$ .

$7 x + 4 - 3 = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Subtract $3$ from $4$ .

$7 x + 1 = \frac{1}{12 (x - 1)} (12 (x - 1) (x + 1))$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$7 x + 1 = 12 \frac{1}{12 (x - 1)} ((x - 1) (x + 1))$

Cancel the common factor of $12$ .

Cancel the common factor.

$7 x + 1 = 12 \frac{1}{12 (x - 1)} ((x - 1) (x + 1))$

Rewrite the expression.

$7 x + 1 = \frac{1}{x - 1} ((x - 1) (x + 1))$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$7 x + 1 = \frac{1}{x - 1} ((x - 1) (x + 1))$

Rewrite the expression.

$7 x + 1 = x + 1$

Step 4

Solve the equation.

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

Subtract $x$ from both sides of the equation.

$7 x + 1 - x = 1$

Subtract $x$ from $7 x$ .

$6 x + 1 = 1$

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

Subtract $1$ from both sides of the equation.

$6 x = 1 - 1$

Subtract $1$ from $1$ .

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x = 0$

Solve for x 1/(3x-3)+1/(4x+4)=1/(12x-12)

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