Solve for a (2x+5)/((x-8)(2x+1))=a/(x-8)+b/(2x+1)

$\frac{2 x + 5}{(x - 8) (2 x + 1)} = \frac{a}{x - 8} + \frac{b}{2 x + 1}$

Step 1

Rewrite the equation as $\frac{a}{x - 8} + \frac{b}{2 x + 1} = \frac{2 x + 5}{(x - 8) (2 x + 1)}$ .

$\frac{a}{x - 8} + \frac{b}{2 x + 1} = \frac{2 x + 5}{(x - 8) (2 x + 1)}$

Step 2

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

Subtract $\frac{2 x + 5}{(x - 8) (2 x + 1)}$ from both sides of the equation.

$\frac{a}{x - 8} + \frac{b}{2 x + 1} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

To write $\frac{a}{x - 8}$ as a fraction with a common denominator, multiply by $\frac{2 x + 1}{2 x + 1}$ .

$\frac{a}{x - 8} \cdot \frac{2 x + 1}{2 x + 1} + \frac{b}{2 x + 1} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

To write $\frac{b}{2 x + 1}$ as a fraction with a common denominator, multiply by $\frac{x - 8}{x - 8}$ .

$\frac{a}{x - 8} \cdot \frac{2 x + 1}{2 x + 1} + \frac{b}{2 x + 1} \cdot \frac{x - 8}{x - 8} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Write each expression with a common denominator of $(x - 8) (2 x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{a}{x - 8}$ by $\frac{2 x + 1}{2 x + 1}$ .

$\frac{a (2 x + 1)}{(x - 8) (2 x + 1)} + \frac{b}{2 x + 1} \cdot \frac{x - 8}{x - 8} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Multiply $\frac{b}{2 x + 1}$ by $\frac{x - 8}{x - 8}$ .

$\frac{a (2 x + 1)}{(x - 8) (2 x + 1)} + \frac{b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Reorder the factors of $(x - 8) (2 x + 1)$ .

$\frac{a (2 x + 1)}{(2 x + 1) (x - 8)} + \frac{b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Combine the numerators over the common denominator.

$\frac{a (2 x + 1) + b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{a (2 x) + a \cdot 1 + b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Rewrite using the commutative property of multiplication.

$\frac{2 a x + a \cdot 1 + b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Multiply $a$ by $1$ .

$\frac{2 a x + a + b (x - 8)}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Apply the distributive property.

$\frac{2 a x + a + b x + b \cdot - 8}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Move $- 8$ to the left of $b$ .

$\frac{2 a x + a + b x - 8 b}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(x - 8) (2 x + 1)} = 0$

Reorder the factors of $(x - 8) (2 x + 1)$ .

$\frac{2 a x + a + b x - 8 b}{(2 x + 1) (x - 8)} - \frac{2 x + 5}{(2 x + 1) (x - 8)} = 0$

Combine the numerators over the common denominator.

$\frac{2 a x + a + b x - 8 b - (2 x + 5)}{(2 x + 1) (x - 8)} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{2 a x + a + b x - 8 b - (2 x) - 1 \cdot 5}{(2 x + 1) (x - 8)} = 0$

Multiply $2$ by $- 1$ .

$\frac{2 a x + a + b x - 8 b - 2 x - 1 \cdot 5}{(2 x + 1) (x - 8)} = 0$

Multiply $- 1$ by $5$ .

$\frac{2 a x + a + b x - 8 b - 2 x - 5}{(2 x + 1) (x - 8)} = 0$

Step 3

Set the numerator equal to zero.

$2 a x + a + b x - 8 b - 2 x - 5 = 0$

Step 4

Solve the equation for $a$ .

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

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

$2 a x + a - 8 b - 2 x - 5 = - b x$

Add $8 b$ to both sides of the equation.

$2 a x + a - 2 x - 5 = - b x + 8 b$

Add $2 x$ to both sides of the equation.

$2 a x + a - 5 = - b x + 8 b + 2 x$

Add $5$ to both sides of the equation.

$2 a x + a = - b x + 8 b + 2 x + 5$

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

Factor $a$ out of $2 a x$ .

$a (2 x) + a = - b x + 8 b + 2 x + 5$

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

$a (2 x) + a = - b x + 8 b + 2 x + 5$

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

$a (2 x) + a \cdot 1 = - b x + 8 b + 2 x + 5$

Factor $a$ out of $a (2 x) + a \cdot 1$ .

$a (2 x + 1) = - b x + 8 b + 2 x + 5$

Divide each term in $a (2 x + 1) = - b x + 8 b + 2 x + 5$ by $2 x + 1$ and simplify.

Divide each term in $a (2 x + 1) = - b x + 8 b + 2 x + 5$ by $2 x + 1$ .

$\frac{a (2 x + 1)}{2 x + 1} = \frac{- b x}{2 x + 1} + \frac{8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Simplify the left side.

Cancel the common factor of $2 x + 1$ .

Cancel the common factor.

$\frac{a (2 x + 1)}{2 x + 1} = \frac{- b x}{2 x + 1} + \frac{8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Divide $a$ by $1$ .

$a = \frac{- b x}{2 x + 1} + \frac{8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Simplify the right side.

Combine into one fraction.

Move the negative in front of the fraction.

$a = - \frac{b x}{2 x + 1} + \frac{8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Combine the numerators over the common denominator.

$a = \frac{- b x + 8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Simplify the numerator.

Factor $b$ out of $- b x + 8 b$ .

Factor $b$ out of $- b x$ .

$a = \frac{b (- 1 x) + 8 b}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Factor $b$ out of $8 b$ .

$a = \frac{b (- 1 x) + b \cdot 8}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Factor $b$ out of $b (- 1 x) + b \cdot 8$ .

$a = \frac{b (- 1 x + 8)}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Rewrite $- 1 x$ as $- x$ .

$a = \frac{b (- x + 8)}{2 x + 1} + \frac{2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Combine the numerators over the common denominator.

$a = \frac{b (- x + 8) + 2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Simplify the numerator.

Apply the distributive property.

$a = \frac{b (- x) + b \cdot 8 + 2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Rewrite using the commutative property of multiplication.

$a = \frac{- b x + b \cdot 8 + 2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Move $8$ to the left of $b$ .

$a = \frac{- b x + 8 b + 2 x}{2 x + 1} + \frac{5}{2 x + 1}$

Simplify terms.

Combine the numerators over the common denominator.

$a = \frac{- b x + 8 b + 2 x + 5}{2 x + 1}$

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

$a = \frac{- (b x) + 8 b + 2 x + 5}{2 x + 1}$

Factor $- 1$ out of $8 b$ .

$a = \frac{- (b x) - (- 8 b) + 2 x + 5}{2 x + 1}$

Factor $- 1$ out of $- (b x) - (- 8 b)$ .

$a = \frac{- (b x - 8 b) + 2 x + 5}{2 x + 1}$

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

$a = \frac{- (b x - 8 b) - (- 2 x) + 5}{2 x + 1}$

Factor $- 1$ out of $- (b x - 8 b) - (- 2 x)$ .

$a = \frac{- (b x - 8 b - 2 x) + 5}{2 x + 1}$

Rewrite $5$ as $- 1 (- 5)$ .

$a = \frac{- (b x - 8 b - 2 x) - 1 (- 5)}{2 x + 1}$

Factor $- 1$ out of $- (b x - 8 b - 2 x) - 1 (- 5)$ .

$a = \frac{- (b x - 8 b - 2 x - 5)}{2 x + 1}$

Simplify the expression.

Rewrite $- (b x - 8 b - 2 x - 5)$ as $- 1 (b x - 8 b - 2 x - 5)$ .

$a = \frac{- 1 (b x - 8 b - 2 x - 5)}{2 x + 1}$

Move the negative in front of the fraction.

$a = - \frac{b x - 8 b - 2 x - 5}{2 x + 1}$

Solve for a (2x+5)/((x-8)(2x+1))=a/(x-8)+b/(2x+1)

Related Questions

Support LCMGCF.com by disabling your adblocker.