Solve x-3+6/x/1-frac{1{x}}

$$\frac{ x-3+ \frac{ 6 }{ x } }{ 1- \frac{ 1 }{ x } }$$

Evaluate

$$\frac{x^{2}-3x+6}{x-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-3 times \frac{x}{x}.

$$\frac{\frac{\left(x-3\right)x}{x}+\frac{6}{x}}{1-\frac{1}{x}}$$

Since \frac{\left(x-3\right)x}{x} and \frac{6}{x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-3\right)x+6}{x}}{1-\frac{1}{x}}$$

Do the multiplications in \left(x-3\right)x+6.

$$\frac{\frac{x^{2}-3x+6}{x}}{1-\frac{1}{x}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{\frac{x^{2}-3x+6}{x}}{\frac{x}{x}-\frac{1}{x}}$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}-3x+6}{x}}{\frac{x-1}{x}}$$

Divide \frac{x^{2}-3x+6}{x} by \frac{x-1}{x} by multiplying \frac{x^{2}-3x+6}{x} by the reciprocal of \frac{x-1}{x}.

$$\frac{\left(x^{2}-3x+6\right)x}{x\left(x-1\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}-3x+6}{x-1}$$

Expand

$$\frac{x^{2}-3x+6}{x-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x-3 times \frac{x}{x}.

$$\frac{\frac{\left(x-3\right)x}{x}+\frac{6}{x}}{1-\frac{1}{x}}$$

Since \frac{\left(x-3\right)x}{x} and \frac{6}{x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-3\right)x+6}{x}}{1-\frac{1}{x}}$$

Do the multiplications in \left(x-3\right)x+6.

$$\frac{\frac{x^{2}-3x+6}{x}}{1-\frac{1}{x}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{\frac{x^{2}-3x+6}{x}}{\frac{x}{x}-\frac{1}{x}}$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x^{2}-3x+6}{x}}{\frac{x-1}{x}}$$

Divide \frac{x^{2}-3x+6}{x} by \frac{x-1}{x} by multiplying \frac{x^{2}-3x+6}{x} by the reciprocal of \frac{x-1}{x}.

$$\frac{\left(x^{2}-3x+6\right)x}{x\left(x-1\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}-3x+6}{x-1}$$