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

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

Evaluate

$$\frac{24}{x+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and x+3 is \left(x-3\right)\left(x+3\right). Multiply \frac{2}{x-3} times \frac{x+3}{x+3}. Multiply \frac{2}{x+3} times \frac{x-3}{x-3}.

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

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

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

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

$$\frac{\frac{2x+6-2x+6}{\left(x-3\right)\left(x+3\right)}+\frac{4}{x^{2}-9}}{\frac{2}{3x-9}}$$

Combine like terms in 2x+6-2x+6.

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

Factor x^{2}-9.

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

Since \frac{12}{\left(x-3\right)\left(x+3\right)} and \frac{4}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators. Add 12 and 4 to get 16.

$$\frac{\frac{16}{\left(x-3\right)\left(x+3\right)}}{\frac{2}{3x-9}}$$

Divide \frac{16}{\left(x-3\right)\left(x+3\right)} by \frac{2}{3x-9} by multiplying \frac{16}{\left(x-3\right)\left(x+3\right)} by the reciprocal of \frac{2}{3x-9}.

$$\frac{16\left(3x-9\right)}{\left(x-3\right)\left(x+3\right)\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{8\left(3x-9\right)}{\left(x-3\right)\left(x+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{3\times 8\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}$$

Cancel out x-3 in both numerator and denominator.

$$\frac{3\times 8}{x+3}$$

Expand the expression.

$$\frac{24}{x+3}$$

Expand

$$\frac{24}{x+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and x+3 is \left(x-3\right)\left(x+3\right). Multiply \frac{2}{x-3} times \frac{x+3}{x+3}. Multiply \frac{2}{x+3} times \frac{x-3}{x-3}.

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

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

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

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

$$\frac{\frac{2x+6-2x+6}{\left(x-3\right)\left(x+3\right)}+\frac{4}{x^{2}-9}}{\frac{2}{3x-9}}$$

Combine like terms in 2x+6-2x+6.

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

Factor x^{2}-9.

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

Since \frac{12}{\left(x-3\right)\left(x+3\right)} and \frac{4}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators. Add 12 and 4 to get 16.

$$\frac{\frac{16}{\left(x-3\right)\left(x+3\right)}}{\frac{2}{3x-9}}$$

Divide \frac{16}{\left(x-3\right)\left(x+3\right)} by \frac{2}{3x-9} by multiplying \frac{16}{\left(x-3\right)\left(x+3\right)} by the reciprocal of \frac{2}{3x-9}.

$$\frac{16\left(3x-9\right)}{\left(x-3\right)\left(x+3\right)\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{8\left(3x-9\right)}{\left(x-3\right)\left(x+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{3\times 8\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}$$

Cancel out x-3 in both numerator and denominator.

$$\frac{3\times 8}{x+3}$$

Expand the expression.

$$\frac{24}{x+3}$$