Solve (1/t^2-4+1/t-2):3t+9/t+2

$$( \frac { 1 } { t ^ { 2 } - 4 } + \frac { 1 } { t - 2 } ) : \frac { 3 t + 9 } { t + 2 }$$

Evaluate

$$\frac{1}{3\left(t-2\right)}$$

Short Solution Steps

Factor t^{2}-4.

$$\frac{\frac{1}{\left(t-2\right)\left(t+2\right)}+\frac{1}{t-2}}{\frac{3t+9}{t+2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(t-2\right)\left(t+2\right) and t-2 is \left(t-2\right)\left(t+2\right). Multiply \frac{1}{t-2} times \frac{t+2}{t+2}.

$$\frac{\frac{1}{\left(t-2\right)\left(t+2\right)}+\frac{t+2}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Since \frac{1}{\left(t-2\right)\left(t+2\right)} and \frac{t+2}{\left(t-2\right)\left(t+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{1+t+2}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Combine like terms in 1+t+2.

$$\frac{\frac{3+t}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Divide \frac{3+t}{\left(t-2\right)\left(t+2\right)} by \frac{3t+9}{t+2} by multiplying \frac{3+t}{\left(t-2\right)\left(t+2\right)} by the reciprocal of \frac{3t+9}{t+2}.

$$\frac{\left(3+t\right)\left(t+2\right)}{\left(t-2\right)\left(t+2\right)\left(3t+9\right)}$$

Cancel out t+2 in both numerator and denominator.

$$\frac{t+3}{\left(t-2\right)\left(3t+9\right)}$$

Factor the expressions that are not already factored.

$$\frac{t+3}{3\left(t-2\right)\left(t+3\right)}$$

Cancel out t+3 in both numerator and denominator.

$$\frac{1}{3\left(t-2\right)}$$

Expand the expression.

$$\frac{1}{3t-6}$$

Expand

$$\frac{1}{3\left(t-2\right)}$$

Short Solution Steps

Factor t^{2}-4.

$$\frac{\frac{1}{\left(t-2\right)\left(t+2\right)}+\frac{1}{t-2}}{\frac{3t+9}{t+2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(t-2\right)\left(t+2\right) and t-2 is \left(t-2\right)\left(t+2\right). Multiply \frac{1}{t-2} times \frac{t+2}{t+2}.

$$\frac{\frac{1}{\left(t-2\right)\left(t+2\right)}+\frac{t+2}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Since \frac{1}{\left(t-2\right)\left(t+2\right)} and \frac{t+2}{\left(t-2\right)\left(t+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{1+t+2}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Combine like terms in 1+t+2.

$$\frac{\frac{3+t}{\left(t-2\right)\left(t+2\right)}}{\frac{3t+9}{t+2}}$$

Divide \frac{3+t}{\left(t-2\right)\left(t+2\right)} by \frac{3t+9}{t+2} by multiplying \frac{3+t}{\left(t-2\right)\left(t+2\right)} by the reciprocal of \frac{3t+9}{t+2}.

$$\frac{\left(3+t\right)\left(t+2\right)}{\left(t-2\right)\left(t+2\right)\left(3t+9\right)}$$

Cancel out t+2 in both numerator and denominator.

$$\frac{t+3}{\left(t-2\right)\left(3t+9\right)}$$

Factor the expressions that are not already factored.

$$\frac{t+3}{3\left(t-2\right)\left(t+3\right)}$$

Cancel out t+3 in both numerator and denominator.

$$\frac{1}{3\left(t-2\right)}$$

Expand the expression.

$$\frac{1}{3t-6}$$