Precalculus Examples

$\frac{4 - y}{(y + 4) (y + 2)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{y + 4}$

Step 1.2

$\frac{A}{y + 4} + \frac{B}{y + 2}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(y + 4) (y + 2)$ .

$\frac{(4 - y) (y + 4) (y + 2)}{(y + 4) (y + 2)} = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.4

Cancel the common factor of $y + 4$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(4 - y) (y + 4) (y + 2)}{(y + 4) (y + 2)} = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.4.2

Rewrite the expression.

$\frac{(4 - y) (y + 2)}{y + 2} = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.5

Cancel the common factor of $y + 2$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(4 - y) (y + 2)}{y + 2} = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.5.2

Divide $4 - y$ by $1$ .

$4 - y = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.6

Reorder $4$ and $- y$ .

$- y + 4 = \frac{(A) (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $y + 4$ .

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Step 1.7.1.1

Cancel the common factor.

$- y + 4 = \frac{A (y + 4) (y + 2)}{y + 4} + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7.1.2

Divide $(A) (y + 2)$ by $1$ .

$- y + 4 = (A) (y + 2) + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7.2

Apply the distributive property.

$- y + 4 = A y + A \cdot 2 + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7.3

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

$- y + 4 = A y + 2 \cdot A + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7.4

Cancel the common factor of $y + 2$ .

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Step 1.7.4.1

Cancel the common factor.

$- y + 4 = A y + 2 A + \frac{(B) (y + 4) (y + 2)}{y + 2}$

Step 1.7.4.2

Divide $(B) (y + 4)$ by $1$ .

$- y + 4 = A y + 2 A + (B) (y + 4)$

Step 1.7.5

Apply the distributive property.

$- y + 4 = A y + 2 A + B y + B \cdot 4$

Step 1.7.6

Move $4$ to the left of $B$ .

$- y + 4 = A y + 2 A + B y + 4 B$

Step 1.8

Move $2 A$ .

$- y + 4 = A y + B y + 2 A + 4 B$

Step 2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.1

Create an equation for the partial fraction variables by equating the coefficients of $y$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$- 1 = A + B$

Step 2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $y$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$4 = 2 A + 4 B$

Step 2.3

Set up the system of equations to find the coefficients of the partial fractions.

$- 1 = A + B$

$4 = 2 A + 4 B$

$- 1 = A + B$

$4 = 2 A + 4 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $- 1 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = - 1$ .

$A + B = - 1$

$4 = 2 A + 4 B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = - 1 - B$

$4 = 2 A + 4 B$

$A = - 1 - B$

$4 = 2 A + 4 B$

Step 3.2

Replace all occurrences of $A$ with $- 1 - B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $4 = 2 A + 4 B$ with $- 1 - B$ .

$4 = 2 (- 1 - B) + 4 B$

$A = - 1 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (- 1 - B) + 4 B$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Apply the distributive property.

$4 = 2 \cdot - 1 + 2 (- B) + 4 B$

$A = - 1 - B$

Step 3.2.2.1.1.2

Multiply $2$ by $- 1$ .

$4 = - 2 + 2 (- B) + 4 B$

$A = - 1 - B$

Step 3.2.2.1.1.3

Multiply $- 1$ by $2$ .

$4 = - 2 - 2 B + 4 B$

$A = - 1 - B$

$4 = - 2 - 2 B + 4 B$

$A = - 1 - B$

Step 3.2.2.1.2

Add $- 2 B$ and $4 B$ .

$4 = - 2 + 2 B$

$A = - 1 - B$

$4 = - 2 + 2 B$

$A = - 1 - B$

$4 = - 2 + 2 B$

$A = - 1 - B$

$4 = - 2 + 2 B$

$A = - 1 - B$

Step 3.3

Solve for $B$ in $4 = - 2 + 2 B$ .

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Step 3.3.1

Rewrite the equation as $- 2 + 2 B = 4$ .

$- 2 + 2 B = 4$

$A = - 1 - B$

Step 3.3.2

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

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Step 3.3.2.1

Add $2$ to both sides of the equation.

$2 B = 4 + 2$

$A = - 1 - B$

Step 3.3.2.2

Add $4$ and $2$ .

$2 B = 6$

$A = - 1 - B$

$2 B = 6$

$A = - 1 - B$

Step 3.3.3

Divide each term in $2 B = 6$ by $2$ and simplify.

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Step 3.3.3.1

Divide each term in $2 B = 6$ by $2$ .

$\frac{2 B}{2} = \frac{6}{2}$

$A = - 1 - B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{6}{2}$

$A = - 1 - B$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{6}{2}$

$A = - 1 - B$

$B = \frac{6}{2}$

$A = - 1 - B$

$B = \frac{6}{2}$

$A = - 1 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $6$ by $2$ .

$B = 3$

$A = - 1 - B$

$B = 3$

$A = - 1 - B$

$B = 3$

$A = - 1 - B$

$B = 3$

$A = - 1 - B$

Step 3.4

Replace all occurrences of $B$ with $3$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = - 1 - B$ with $3$ .

$A = - 1 - (3)$

$B = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 1 - (3)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $3$ .

$A = - 1 - 3$

$B = 3$

Step 3.4.2.1.2

Subtract $3$ from $- 1$ .

$A = - 4$

$B = 3$

$A = - 4$

$B = 3$

$A = - 4$

$B = 3$

$A = - 4$

$B = 3$

Step 3.5

List all of the solutions.

$A = - 4, B = 3$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{y + 4} + \frac{B}{y + 2}$ with the values found for $A$ and $B$ .

$\frac{- 4}{y + 4} + \frac{3}{y + 2}$

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