Solve d-2/6*(14+d^2/d^2-4-d-4/d+2)

$$\frac { d - 2 } { 6 } \cdot ( \frac { 14 + d ^ { 2 } } { d ^ { 2 } - 4 } - \frac { d - 4 } { d + 2 } )$$

Evaluate

$$\frac{d+1}{d+2}$$

Short Solution Steps

Factor d^{2}-4.

$$\frac{d-2}{6}\left(\frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)}-\frac{d-4}{d+2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(d-2\right)\left(d+2\right) and d+2 is \left(d-2\right)\left(d+2\right). Multiply \frac{d-4}{d+2} times \frac{d-2}{d-2}.

$$\frac{d-2}{6}\left(\frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)}-\frac{\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)}\right)$$

Since \frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)} and \frac{\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{d-2}{6}\times \frac{14+d^{2}-\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)}$$

Do the multiplications in 14+d^{2}-\left(d-4\right)\left(d-2\right).

$$\frac{d-2}{6}\times \frac{14+d^{2}-d^{2}+2d+4d-8}{\left(d-2\right)\left(d+2\right)}$$

Combine like terms in 14+d^{2}-d^{2}+2d+4d-8.

$$\frac{d-2}{6}\times \frac{6+6d}{\left(d-2\right)\left(d+2\right)}$$

Multiply \frac{d-2}{6} times \frac{6+6d}{\left(d-2\right)\left(d+2\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(d-2\right)\left(6+6d\right)}{6\left(d-2\right)\left(d+2\right)}$$

Cancel out d-2 in both numerator and denominator.

$$\frac{6d+6}{6\left(d+2\right)}$$

Factor the expressions that are not already factored.

$$\frac{6\left(d+1\right)}{6\left(d+2\right)}$$

Cancel out 6 in both numerator and denominator.

$$\frac{d+1}{d+2}$$

Expand

$$\frac{d+1}{d+2}$$

Short Solution Steps

Factor d^{2}-4.

$$\frac{d-2}{6}\left(\frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)}-\frac{d-4}{d+2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(d-2\right)\left(d+2\right) and d+2 is \left(d-2\right)\left(d+2\right). Multiply \frac{d-4}{d+2} times \frac{d-2}{d-2}.

$$\frac{d-2}{6}\left(\frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)}-\frac{\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)}\right)$$

Since \frac{14+d^{2}}{\left(d-2\right)\left(d+2\right)} and \frac{\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{d-2}{6}\times \frac{14+d^{2}-\left(d-4\right)\left(d-2\right)}{\left(d-2\right)\left(d+2\right)}$$

Do the multiplications in 14+d^{2}-\left(d-4\right)\left(d-2\right).

$$\frac{d-2}{6}\times \frac{14+d^{2}-d^{2}+2d+4d-8}{\left(d-2\right)\left(d+2\right)}$$

Combine like terms in 14+d^{2}-d^{2}+2d+4d-8.

$$\frac{d-2}{6}\times \frac{6+6d}{\left(d-2\right)\left(d+2\right)}$$

Multiply \frac{d-2}{6} times \frac{6+6d}{\left(d-2\right)\left(d+2\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(d-2\right)\left(6+6d\right)}{6\left(d-2\right)\left(d+2\right)}$$

Cancel out d-2 in both numerator and denominator.

$$\frac{6d+6}{6\left(d+2\right)}$$

Factor the expressions that are not already factored.

$$\frac{6\left(d+1\right)}{6\left(d+2\right)}$$

Cancel out 6 in both numerator and denominator.

$$\frac{d+1}{d+2}$$