Simplify (2b)/(2b+3)+5/(3-2b)-(4b^2+9)/(4b^2-9)

$\frac{2 b}{2 b + 3} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{4 b^{2} - 9}$

Simplify the denominator.

Rewrite $4 b^{2}$ as ${(2 b)}^{2}$ .

$\frac{2 b}{2 b + 3} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{{(2 b)}^{2} - 9}$

Rewrite $9$ as $3^{2}$ .

$\frac{2 b}{2 b + 3} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{{(2 b)}^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 b$ and $b = 3$ .

$\frac{2 b}{2 b + 3} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

To write $\frac{2 b}{2 b + 3}$ as a fraction with a common denominator, multiply by $\frac{3 - 2 b}{3 - 2 b}$ .

$\frac{2 b}{2 b + 3} \cdot \frac{3 - 2 b}{3 - 2 b} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

To write $\frac{5}{3 - 2 b}$ as a fraction with a common denominator, multiply by $\frac{2 b + 3}{2 b + 3}$ .

$\frac{2 b}{2 b + 3} \cdot \frac{3 - 2 b}{3 - 2 b} + \frac{5}{3 - 2 b} \cdot \frac{2 b + 3}{2 b + 3} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Write each expression with a common denominator of $(2 b + 3) (3 - 2 b)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 b}{2 b + 3}$ by $\frac{3 - 2 b}{3 - 2 b}$ .

$\frac{2 b (3 - 2 b)}{(2 b + 3) (3 - 2 b)} + \frac{5}{3 - 2 b} \cdot \frac{2 b + 3}{2 b + 3} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $\frac{5}{3 - 2 b}$ by $\frac{2 b + 3}{2 b + 3}$ .

$\frac{2 b (3 - 2 b)}{(2 b + 3) (3 - 2 b)} + \frac{5 (2 b + 3)}{(3 - 2 b) (2 b + 3)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Reorder the factors of $(3 - 2 b) (2 b + 3)$ .

$\frac{2 b (3 - 2 b)}{(2 b + 3) (3 - 2 b)} + \frac{5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Combine the numerators over the common denominator.

$\frac{2 b (3 - 2 b) + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Simplify the numerator.

Apply the distributive property.

$\frac{2 b \cdot 3 + 2 b (- 2 b) + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $3$ by $2$ .

$\frac{6 b + 2 b (- 2 b) + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Rewrite using the commutative property of multiplication.

$\frac{6 b + 2 \cdot - 2 b \cdot b + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Simplify each term.

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$\frac{6 b + 2 \cdot - 2 (b \cdot b) + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $b$ by $b$ .

$\frac{6 b + 2 \cdot - 2 b^{2} + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $2$ by $- 2$ .

$\frac{6 b - 4 b^{2} + 5 (2 b + 3)}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Apply the distributive property.

$\frac{6 b - 4 b^{2} + 5 (2 b) + 5 \cdot 3}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $2$ by $5$ .

$\frac{6 b - 4 b^{2} + 10 b + 5 \cdot 3}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Multiply $5$ by $3$ .

$\frac{6 b - 4 b^{2} + 10 b + 15}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Add $6 b$ and $10 b$ .

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (2 b - 3)}$

Simplify with factoring out.

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

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b) - 3)}$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b) - 1 (3))}$

Factor $- 1$ out of $- (- 2 b) - 1 (3)$ .

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (3 - 2 b)} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b + 3))}$

Reorder terms.

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (- 2 b + 3)} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b + 3))}$

To write $\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (- 2 b + 3)}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (- 2 b + 3)} \cdot \frac{- 1}{- 1} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b + 3))}$

Write each expression with a common denominator of $(2 b + 3) (- 2 b + 3) \cdot - 1$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{- 4 b^{2} + 16 b + 15}{(2 b + 3) (- 2 b + 3)}$ by $\frac{- 1}{- 1}$ .

$\frac{(- 4 b^{2} + 16 b + 15) \cdot - 1}{(2 b + 3) (- 2 b + 3) \cdot - 1} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b + 3))}$

Reorder the factors of $(2 b + 3) (- 2 b + 3) \cdot - 1$ .

$\frac{(- 4 b^{2} + 16 b + 15) \cdot - 1}{- (- 2 b + 3) (2 b + 3)} - \frac{4 b^{2} + 9}{(2 b + 3) (- (- 2 b + 3))}$

Reorder the factors of $(2 b + 3) (- (- 2 b + 3))$ .

$\frac{(- 4 b^{2} + 16 b + 15) \cdot - 1}{- (- 2 b + 3) (2 b + 3)} - \frac{4 b^{2} + 9}{- (- 2 b + 3) (2 b + 3)}$

Combine the numerators over the common denominator.

$\frac{(- 4 b^{2} + 16 b + 15) \cdot - 1 - (4 b^{2} + 9)}{- (- 2 b + 3) (2 b + 3)}$

Simplify the numerator.

Apply the distributive property.

$\frac{- 4 b^{2} \cdot - 1 + 16 b \cdot - 1 + 15 \cdot - 1 - (4 b^{2} + 9)}{- (- 2 b + 3) (2 b + 3)}$

Simplify.

Multiply $- 1$ by $- 4$ .

$\frac{4 b^{2} + 16 b \cdot - 1 + 15 \cdot - 1 - (4 b^{2} + 9)}{- (- 2 b + 3) (2 b + 3)}$

Multiply $- 1$ by $16$ .

$\frac{4 b^{2} - 16 b + 15 \cdot - 1 - (4 b^{2} + 9)}{- (- 2 b + 3) (2 b + 3)}$

Multiply $15$ by $- 1$ .

$\frac{4 b^{2} - 16 b - 15 - (4 b^{2} + 9)}{- (- 2 b + 3) (2 b + 3)}$

Apply the distributive property.

$\frac{4 b^{2} - 16 b - 15 - (4 b^{2}) - 1 \cdot 9}{- (- 2 b + 3) (2 b + 3)}$

Multiply $4$ by $- 1$ .

$\frac{4 b^{2} - 16 b - 15 - 4 b^{2} - 1 \cdot 9}{- (- 2 b + 3) (2 b + 3)}$

Multiply $- 1$ by $9$ .

$\frac{4 b^{2} - 16 b - 15 - 4 b^{2} - 9}{- (- 2 b + 3) (2 b + 3)}$

Subtract $4 b^{2}$ from $4 b^{2}$ .

$\frac{- 16 b - 15 + 0 - 9}{- (- 2 b + 3) (2 b + 3)}$

Add $- 16 b$ and $0$ .

$\frac{- 16 b - 15 - 9}{- (- 2 b + 3) (2 b + 3)}$

Subtract $9$ from $- 15$ .

$\frac{- 16 b - 24}{- (- 2 b + 3) (2 b + 3)}$

Factor $8$ out of $- 16 b - 24$ .

Factor $8$ out of $- 16 b$ .

$\frac{8 (- 2 b) - 24}{- (- 2 b + 3) (2 b + 3)}$

Factor $8$ out of $- 24$ .

$\frac{8 (- 2 b) + 8 (- 3)}{- (- 2 b + 3) (2 b + 3)}$

Factor $8$ out of $8 (- 2 b) + 8 (- 3)$ .

$\frac{8 (- 2 b - 3)}{- (- 2 b + 3) (2 b + 3)}$

Simplify terms.

Cancel the common factor of $- 2 b - 3$ and $2 b + 3$ .

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

$\frac{8 (- (2 b) - 3)}{- (- 2 b + 3) (2 b + 3)}$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{8 (- (2 b) - 1 (3))}{- (- 2 b + 3) (2 b + 3)}$

Factor $- 1$ out of $- (2 b) - 1 (3)$ .

$\frac{8 (- (2 b + 3))}{- (- 2 b + 3) (2 b + 3)}$

Rewrite $- (2 b + 3)$ as $- 1 (2 b + 3)$ .

$\frac{8 (- 1 (2 b + 3))}{- (- 2 b + 3) (2 b + 3)}$

Cancel the common factor.

$\frac{8 (- 1 (2 b + 3))}{- (- 2 b + 3) (2 b + 3)}$

Rewrite the expression.

$\frac{8 \cdot (- 1)}{- (- 2 b + 3)}$

Dividing two negative values results in a positive value.

$\frac{8}{- 2 b + 3}$

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

$\frac{8}{- (2 b) + 3}$

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{8}{- (2 b) - 1 (- 3)}$

Factor $- 1$ out of $- (2 b) - 1 (- 3)$ .

$\frac{8}{- (2 b - 3)}$

Rewrite negatives.

Rewrite $- (2 b - 3)$ as $- 1 (2 b - 3)$ .

$\frac{8}{- 1 (2 b - 3)}$

Move the negative in front of the fraction.

$- \frac{8}{2 b - 3}$

Simplify (2b)/(2b+3)+5/(3-2b)-(4b^2+9)/(4b^2-9)

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