Solve for x (2x+4)/(x^2+4x+3)=1/(x+1)+1/(x+3)

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

Factor each term.

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $4$ .

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

Factor $2$ out of $2 x + 2 \cdot 2$ .

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

Factor $x^{2} + 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

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.

$(x + 1) (x + 3), x + 1, x + 3$

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

$1$

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

$(x + 1) = x + 1$

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

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

$(x + 3) = x + 3$

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

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

$(x + 1) = x + 1$

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

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

$(x + 3) = x + 3$

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

The LCM of $x + 1, x + 3, x + 1, x + 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 1) (x + 3)$

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

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

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

Simplify the left side.

Cancel the common factor of $(x + 1) (x + 3)$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

$2 x + 4 = x + 3 + x + 1$

Simplify by adding terms.

Add $x$ and $x$ .

$2 x + 4 = 2 x + 3 + 1$

Add $3$ and $1$ .

$2 x + 4 = 2 x + 4$

Solve the equation.

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

Subtract $2 x$ from both sides of the equation.

$2 x + 4 - 2 x = 4$

Combine the opposite terms in $2 x + 4 - 2 x$ .

Subtract $2 x$ from $2 x$ .

$0 + 4 = 4$

Add $0$ and $4$ .

$4 = 4$

Since $4 = 4$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x (2x+4)/(x^2+4x+3)=1/(x+1)+1/(x+3)

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