Solve x^2-x/3x+9/x^2-2x+1/18-2x^2

$$\frac { x ^ { 2 } - x } { 3 x + 9 } / \frac { x ^ { 2 } - 2 x + 1 } { 18 - 2 x ^ { 2 } }$$

Evaluate

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

Short Solution Steps

Divide \frac{x^{2}-x}{3x+9} by \frac{x^{2}-2x+1}{18-2x^{2}} by multiplying \frac{x^{2}-x}{3x+9} by the reciprocal of \frac{x^{2}-2x+1}{18-2x^{2}}.

$$\frac{\left(x^{2}-x\right)\left(18-2x^{2}\right)}{\left(3x+9\right)\left(x^{2}-2x+1\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in -3-x.

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

Cancel out \left(x-1\right)\left(x+3\right) in both numerator and denominator.

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

Expand the expression.

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

Expand

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

Short Solution Steps

Divide \frac{x^{2}-x}{3x+9} by \frac{x^{2}-2x+1}{18-2x^{2}} by multiplying \frac{x^{2}-x}{3x+9} by the reciprocal of \frac{x^{2}-2x+1}{18-2x^{2}}.

$$\frac{\left(x^{2}-x\right)\left(18-2x^{2}\right)}{\left(3x+9\right)\left(x^{2}-2x+1\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in -3-x.

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

Cancel out \left(x-1\right)\left(x+3\right) in both numerator and denominator.

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

Expand the expression.

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