Solve x-4+9/2x+3/x+3-frac{5{2x+3}}

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

Evaluate

$$\frac{x-3}{x+4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-4 times \frac{2x+3}{2x+3}.

$$\frac{\frac{\left(x-4\right)\left(2x+3\right)}{2x+3}+\frac{9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Since \frac{\left(x-4\right)\left(2x+3\right)}{2x+3} and \frac{9}{2x+3} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-4\right)\left(2x+3\right)+9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Do the multiplications in \left(x-4\right)\left(2x+3\right)+9.

$$\frac{\frac{2x^{2}+3x-8x-12+9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Combine like terms in 2x^{2}+3x-8x-12+9.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{x+3-\frac{5}{2x+3}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x+3 times \frac{2x+3}{2x+3}.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{\left(x+3\right)\left(2x+3\right)}{2x+3}-\frac{5}{2x+3}}$$

Since \frac{\left(x+3\right)\left(2x+3\right)}{2x+3} and \frac{5}{2x+3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{\left(x+3\right)\left(2x+3\right)-5}{2x+3}}$$

Do the multiplications in \left(x+3\right)\left(2x+3\right)-5.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{2x^{2}+3x+6x+9-5}{2x+3}}$$

Combine like terms in 2x^{2}+3x+6x+9-5.

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

Divide \frac{2x^{2}-5x-3}{2x+3} by \frac{2x^{2}+9x+4}{2x+3} by multiplying \frac{2x^{2}-5x-3}{2x+3} by the reciprocal of \frac{2x^{2}+9x+4}{2x+3}.

$$\frac{\left(2x^{2}-5x-3\right)\left(2x+3\right)}{\left(2x+3\right)\left(2x^{2}+9x+4\right)}$$

Cancel out 2x+3 in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{\left(x-3\right)\left(2x+1\right)}{\left(x+4\right)\left(2x+1\right)}$$

Cancel out 2x+1 in both numerator and denominator.

$$\frac{x-3}{x+4}$$

Expand

$$\frac{x-3}{x+4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-4 times \frac{2x+3}{2x+3}.

$$\frac{\frac{\left(x-4\right)\left(2x+3\right)}{2x+3}+\frac{9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Since \frac{\left(x-4\right)\left(2x+3\right)}{2x+3} and \frac{9}{2x+3} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-4\right)\left(2x+3\right)+9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Do the multiplications in \left(x-4\right)\left(2x+3\right)+9.

$$\frac{\frac{2x^{2}+3x-8x-12+9}{2x+3}}{x+3-\frac{5}{2x+3}}$$

Combine like terms in 2x^{2}+3x-8x-12+9.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{x+3-\frac{5}{2x+3}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x+3 times \frac{2x+3}{2x+3}.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{\left(x+3\right)\left(2x+3\right)}{2x+3}-\frac{5}{2x+3}}$$

Since \frac{\left(x+3\right)\left(2x+3\right)}{2x+3} and \frac{5}{2x+3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{\left(x+3\right)\left(2x+3\right)-5}{2x+3}}$$

Do the multiplications in \left(x+3\right)\left(2x+3\right)-5.

$$\frac{\frac{2x^{2}-5x-3}{2x+3}}{\frac{2x^{2}+3x+6x+9-5}{2x+3}}$$

Combine like terms in 2x^{2}+3x+6x+9-5.

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

Divide \frac{2x^{2}-5x-3}{2x+3} by \frac{2x^{2}+9x+4}{2x+3} by multiplying \frac{2x^{2}-5x-3}{2x+3} by the reciprocal of \frac{2x^{2}+9x+4}{2x+3}.

$$\frac{\left(2x^{2}-5x-3\right)\left(2x+3\right)}{\left(2x+3\right)\left(2x^{2}+9x+4\right)}$$

Cancel out 2x+3 in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{\left(x-3\right)\left(2x+1\right)}{\left(x+4\right)\left(2x+1\right)}$$

Cancel out 2x+1 in both numerator and denominator.

$$\frac{x-3}{x+4}$$