Solve 1/x+x/3x+9-3/x^2+3x

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

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Short Solution Steps

Factor 3x+9.

$$\frac{1}{x}+\frac{x}{3\left(x+3\right)}-\frac{3}{x^{2}+3x}$$

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

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

Since \frac{3\left(x+3\right)}{3x\left(x+3\right)} and \frac{xx}{3x\left(x+3\right)} have the same denominator, add them by adding their numerators.

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

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

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

Factor x^{2}+3x.

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

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

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

Since \frac{3x+9+x^{2}}{3x\left(x+3\right)} and \frac{3\times 3}{3x\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3x+9+x^{2}-3\times 3.

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

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

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

Factor the expressions that are not already factored in \frac{3x+x^{2}}{3x\left(x+3\right)}.

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

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

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$