Solve (3/x-x/3):(1/x+1/3)=

$$( \frac { 3 } { x } - \frac { x } { 3 } ) : ( \frac { 1 } { x } + \frac { 1 } { 3 } ) =$$

Evaluate

$$3-x$$

Short Solution Steps

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

$$\frac{\frac{3\times 3}{3x}-\frac{xx}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

Since \frac{3\times 3}{3x} and \frac{xx}{3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3\times 3-xx}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

Do the multiplications in 3\times 3-xx.

$$\frac{\frac{9-x^{2}}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 3 is 3x. Multiply \frac{1}{x} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{x}{x}.

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

Since \frac{3}{3x} and \frac{x}{3x} have the same denominator, add them by adding their numerators.

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

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

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

Cancel out 3x in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in -3-x.

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

Cancel out x+3 in both numerator and denominator.

$$-\left(x-3\right)$$

Expand the expression.

$$-x+3$$

Expand

$$3-x$$

Short Solution Steps

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

$$\frac{\frac{3\times 3}{3x}-\frac{xx}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

Since \frac{3\times 3}{3x} and \frac{xx}{3x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3\times 3-xx}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

Do the multiplications in 3\times 3-xx.

$$\frac{\frac{9-x^{2}}{3x}}{\frac{1}{x}+\frac{1}{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 3 is 3x. Multiply \frac{1}{x} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{x}{x}.

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

Since \frac{3}{3x} and \frac{x}{3x} have the same denominator, add them by adding their numerators.

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

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

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

Cancel out 3x in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in -3-x.

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

Cancel out x+3 in both numerator and denominator.

$$-\left(x-3\right)$$

Expand the expression.

$$-x+3$$