Solve for B 1/((x+1)(x+2))=A/(x+1)+B/(x+2)

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

Rewrite the equation as $\frac{A}{x + 1} + \frac{B}{x + 2} = \frac{1}{(x + 1) (x + 2)}$ .

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

Move all terms containing variables to the left side of the equation.

Subtract $\frac{1}{(x + 1) (x + 2)}$ from both sides of the equation.

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Move $2$ to the left of $A$ .

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

Apply the distributive property.

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

Multiply $B$ by $1$ .

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

Combine the numerators over the common denominator.

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

Set the numerator equal to zero.

$A x + 2 A + B x + B - 1 = 0$

Solve the equation for $B$ .

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

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

$2 A + B x + B - 1 = - A x$

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

$B x + B - 1 = - A x - 2 A$

Add $1$ to both sides of the equation.

$B x + B = - A x - 2 A + 1$

Factor $B$ out of $B x + B$ .

Factor $B$ out of $B x$ .

$B (x) + B = - A x - 2 A + 1$

Raise $B$ to the power of $1$ .

$B (x) + B = - A x - 2 A + 1$

Factor $B$ out of $B^{1}$ .

$B (x) + B \cdot 1 = - A x - 2 A + 1$

Factor $B$ out of $B (x) + B \cdot 1$ .

$B (x + 1) = - A x - 2 A + 1$

Divide each term in $B (x + 1) = - A x - 2 A + 1$ by $x + 1$ and simplify.

Divide each term in $B (x + 1) = - A x - 2 A + 1$ by $x + 1$ .

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

Simplify the left side.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Divide $B$ by $1$ .

$B = \frac{- A x}{x + 1} + \frac{- 2 A}{x + 1} + \frac{1}{x + 1}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$B = - \frac{A x}{x + 1} + \frac{- 2 A}{x + 1} + \frac{1}{x + 1}$

Move the negative in front of the fraction.

$B = - \frac{A x}{x + 1} - \frac{2 A}{x + 1} + \frac{1}{x + 1}$

Simplify terms.

Combine the numerators over the common denominator.

$B = \frac{- A x - 2 A}{x + 1} + \frac{1}{x + 1}$

Combine the numerators over the common denominator.

$B = \frac{- A x - 2 A + 1}{x + 1}$

Factor $- 1$ out of $- A x$ .

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

Factor $- 1$ out of $- 2 A$ .

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

Factor $- 1$ out of $- (A x) - (2 A)$ .

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

Rewrite $1$ as $- 1 (- 1)$ .

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

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

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

Simplify the expression.

Rewrite $- (A x + 2 A - 1)$ as $- 1 (A x + 2 A - 1)$ .

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

Move the negative in front of the fraction.

$B = - \frac{A x + 2 A - 1}{x + 1}$

Solve for B 1/((x+1)(x+2))=A/(x+1)+B/(x+2)

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