Solve x+4/x+3-x-3/x+4/frac{14{x^2+7x+12}}

$$\frac{ \frac{ x+4 }{ x+3 } - \frac{ x-3 }{ x+4 } }{ \frac{ 14 }{ { x }^{ 2 } +7x+12 } }$$

Evaluate

$$\frac{4x}{7}+\frac{25}{14}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+3 and x+4 is \left(x+3\right)\left(x+4\right). Multiply \frac{x+4}{x+3} times \frac{x+4}{x+4}. Multiply \frac{x-3}{x+4} times \frac{x+3}{x+3}.

$$\frac{\frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}-\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

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

$$\frac{\frac{\left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

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

$$\frac{\frac{x^{2}+4x+4x+16-x^{2}-3x+3x+9}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

Combine like terms in x^{2}+4x+4x+16-x^{2}-3x+3x+9.

$$\frac{\frac{8x+25}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

Divide \frac{8x+25}{\left(x+3\right)\left(x+4\right)} by \frac{14}{x^{2}+7x+12} by multiplying \frac{8x+25}{\left(x+3\right)\left(x+4\right)} by the reciprocal of \frac{14}{x^{2}+7x+12}.

$$\frac{\left(8x+25\right)\left(x^{2}+7x+12\right)}{\left(x+3\right)\left(x+4\right)\times 14}$$

Factor the expressions that are not already factored.

$$\frac{\left(x+3\right)\left(x+4\right)\left(8x+25\right)}{14\left(x+3\right)\left(x+4\right)}$$

Cancel out \left(x+3\right)\left(x+4\right) in both numerator and denominator.

$$\frac{8x+25}{14}$$

Expand

$$\frac{4x}{7}+\frac{25}{14}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+3 and x+4 is \left(x+3\right)\left(x+4\right). Multiply \frac{x+4}{x+3} times \frac{x+4}{x+4}. Multiply \frac{x-3}{x+4} times \frac{x+3}{x+3}.

$$\frac{\frac{\left(x+4\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}-\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

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

$$\frac{\frac{\left(x+4\right)\left(x+4\right)-\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

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

$$\frac{\frac{x^{2}+4x+4x+16-x^{2}-3x+3x+9}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

Combine like terms in x^{2}+4x+4x+16-x^{2}-3x+3x+9.

$$\frac{\frac{8x+25}{\left(x+3\right)\left(x+4\right)}}{\frac{14}{x^{2}+7x+12}}$$

Divide \frac{8x+25}{\left(x+3\right)\left(x+4\right)} by \frac{14}{x^{2}+7x+12} by multiplying \frac{8x+25}{\left(x+3\right)\left(x+4\right)} by the reciprocal of \frac{14}{x^{2}+7x+12}.

$$\frac{\left(8x+25\right)\left(x^{2}+7x+12\right)}{\left(x+3\right)\left(x+4\right)\times 14}$$

Factor the expressions that are not already factored.

$$\frac{\left(x+3\right)\left(x+4\right)\left(8x+25\right)}{14\left(x+3\right)\left(x+4\right)}$$

Cancel out \left(x+3\right)\left(x+4\right) in both numerator and denominator.

$$\frac{8x+25}{14}$$