Solve 2x-1/x^2-6x+9-3x/x+3

$$\frac { 2 x - 1 } { x ^ { 2 } - 6 x + 9 } - \frac { 3 x } { x + 3 }$$

Evaluate

$$\frac{-3x^{3}+20x^{2}-22x-3}{\left(x+3\right)\left(x-3\right)^{2}}$$

Short Solution Steps

Factor x^{2}-6x+9.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)^{2} and x+3 is \left(x+3\right)\left(x-3\right)^{2}. Multiply \frac{2x-1}{\left(x-3\right)^{2}} times \frac{x+3}{x+3}. Multiply \frac{3x}{x+3} times \frac{\left(x-3\right)^{2}}{\left(x-3\right)^{2}}.

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

Since \frac{\left(2x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)^{2}} and \frac{3x\left(x-3\right)^{2}}{\left(x+3\right)\left(x-3\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(2x-1\right)\left(x+3\right)-3x\left(x-3\right)^{2}.

$$\frac{2x^{2}+6x-x-3-3x^{3}+18x^{2}-27x}{\left(x+3\right)\left(x-3\right)^{2}}$$

Combine like terms in 2x^{2}+6x-x-3-3x^{3}+18x^{2}-27x.

$$\frac{20x^{2}-22x-3-3x^{3}}{\left(x+3\right)\left(x-3\right)^{2}}$$

Expand \left(x+3\right)\left(x-3\right)^{2}.

$$\frac{20x^{2}-22x-3-3x^{3}}{x^{3}-3x^{2}-9x+27}$$

Expand

$$\frac{-3x^{3}+20x^{2}-22x-3}{\left(x+3\right)\left(x^{2}-6x+9\right)}$$

Short Solution Steps

Factor x^{2}-6x+9.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)^{2} and x+3 is \left(x+3\right)\left(x-3\right)^{2}. Multiply \frac{2x-1}{\left(x-3\right)^{2}} times \frac{x+3}{x+3}. Multiply \frac{3x}{x+3} times \frac{\left(x-3\right)^{2}}{\left(x-3\right)^{2}}.

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

Since \frac{\left(2x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)^{2}} and \frac{3x\left(x-3\right)^{2}}{\left(x+3\right)\left(x-3\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(2x-1\right)\left(x+3\right)-3x\left(x-3\right)^{2}.

$$\frac{2x^{2}+6x-x-3-3x^{3}+18x^{2}-27x}{\left(x+3\right)\left(x-3\right)^{2}}$$

Combine like terms in 2x^{2}+6x-x-3-3x^{3}+18x^{2}-27x.

$$\frac{20x^{2}-22x-3-3x^{3}}{\left(x+3\right)\left(x-3\right)^{2}}$$

Expand \left(x+3\right)\left(x-3\right)^{2}.

$$\frac{20x^{2}-22x-3-3x^{3}}{x^{3}-3x^{2}-9x+27}$$