Algebra Examples

$\frac{(6 x + 5)}{x^{2} - 5 x + 6}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 1.1.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 1.1.2

Write the factored form using these integers.

$\frac{6 x + 5}{(x - 3) (x - 2)}$

Step 1.2

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 $A$ .

$\frac{A}{x - 3}$

Step 1.3

$\frac{A}{x - 3} + \frac{B}{x - 2}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 3) (x - 2)$ .

$\frac{(6 x + 5) (x - 3) (x - 2)}{(x - 3) (x - 2)} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.5

Cancel the common factor of $x - 3$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(6 x + 5) (x - 3) (x - 2)}{(x - 3) (x - 2)} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.5.2

Rewrite the expression.

$\frac{(6 x + 5) (x - 2)}{x - 2} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.6

Cancel the common factor of $x - 2$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(6 x + 5) (x - 2)}{x - 2} = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.6.2

Divide $6 x + 5$ by $1$ .

$6 x + 5 = \frac{(A) (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x - 3$ .

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Step 1.7.1.1

Cancel the common factor.

$6 x + 5 = \frac{A (x - 3) (x - 2)}{x - 3} + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7.1.2

Divide $(A) (x - 2)$ by $1$ .

$6 x + 5 = (A) (x - 2) + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7.2

Apply the distributive property.

$6 x + 5 = A x + A \cdot - 2 + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7.3

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

$6 x + 5 = A x - 2 \cdot A + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7.4

Cancel the common factor of $x - 2$ .

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Step 1.7.4.1

Cancel the common factor.

$6 x + 5 = A x - 2 A + \frac{(B) (x - 3) (x - 2)}{x - 2}$

Step 1.7.4.2

Divide $(B) (x - 3)$ by $1$ .

$6 x + 5 = A x - 2 A + (B) (x - 3)$

Step 1.7.5

Apply the distributive property.

$6 x + 5 = A x - 2 A + B x + B \cdot - 3$

Step 1.7.6

Move $- 3$ to the left of $B$ .

$6 x + 5 = A x - 2 A + B x - 3 B$

Step 1.8

Move $- 2 A$ .

$6 x + 5 = A x + B x - 2 A - 3 B$

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 $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$6 = A + B$

Step 2.2

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

$5 = - 2 A - 3 B$

Step 2.3

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

$6 = A + B$

$5 = - 2 A - 3 B$

$6 = A + B$

$5 = - 2 A - 3 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $6 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 6$ .

$A + B = 6$

$5 = - 2 A - 3 B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 6 - B$

$5 = - 2 A - 3 B$

$A = 6 - B$

$5 = - 2 A - 3 B$

Step 3.2

Replace all occurrences of $A$ with $6 - B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $5 = - 2 A - 3 B$ with $6 - B$ .

$5 = - 2 (6 - B) - 3 B$

$A = 6 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 2 (6 - B) - 3 B$ .

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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.

$5 = - 2 \cdot 6 - 2 (- B) - 3 B$

$A = 6 - B$

Step 3.2.2.1.1.2

Multiply $- 2$ by $6$ .

$5 = - 12 - 2 (- B) - 3 B$

$A = 6 - B$

Step 3.2.2.1.1.3

Multiply $- 1$ by $- 2$ .

$5 = - 12 + 2 B - 3 B$

$A = 6 - B$

$5 = - 12 + 2 B - 3 B$

$A = 6 - B$

Step 3.2.2.1.2

Subtract $3 B$ from $2 B$ .

$5 = - 12 - B$

$A = 6 - B$

$5 = - 12 - B$

$A = 6 - B$

$5 = - 12 - B$

$A = 6 - B$

$5 = - 12 - B$

$A = 6 - B$

Step 3.3

Solve for $B$ in $5 = - 12 - B$ .

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Step 3.3.1

Rewrite the equation as $- 12 - B = 5$ .

$- 12 - B = 5$

$A = 6 - B$

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 $12$ to both sides of the equation.

$- B = 5 + 12$

$A = 6 - B$

Step 3.3.2.2

Add $5$ and $12$ .

$- B = 17$

$A = 6 - B$

$- B = 17$

$A = 6 - B$

Step 3.3.3

Divide each term in $- B = 17$ by $- 1$ and simplify.

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Step 3.3.3.1

Divide each term in $- B = 17$ by $- 1$ .

$\frac{- B}{- 1} = \frac{17}{- 1}$

$A = 6 - B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{17}{- 1}$

$A = 6 - B$

Step 3.3.3.2.2

Divide $B$ by $1$ .

$B = \frac{17}{- 1}$

$A = 6 - B$

$B = \frac{17}{- 1}$

$A = 6 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $17$ by $- 1$ .

$B = - 17$

$A = 6 - B$

$B = - 17$

$A = 6 - B$

$B = - 17$

$A = 6 - B$

$B = - 17$

$A = 6 - B$

Step 3.4

Replace all occurrences of $B$ with $- 17$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = 6 - B$ with $- 17$ .

$A = 6 - (- 17)$

$B = - 17$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $6 - (- 17)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $- 17$ .

$A = 6 + 17$

$B = - 17$

Step 3.4.2.1.2

Add $6$ and $17$ .

$A = 23$

$B = - 17$

$A = 23$

$B = - 17$

$A = 23$

$B = - 17$

$A = 23$

$B = - 17$

Step 3.5

List all of the solutions.

$A = 23, B = - 17$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x - 3} + \frac{B}{x - 2}$ with the values found for $A$ and $B$ .

$\frac{23}{x - 3} + \frac{- 17}{x - 2}$