Solve 1/x^2-3x+2quad1/4x^2-8x

$$\frac { 1 } { x ^ { 2 } - 3 x } + 2 \quad \frac { 1 } { 4 x ^ { 2 } - 8 x }$$

Evaluate

$$\frac{3x-7}{2x\left(x-3\right)\left(x-2\right)}$$

Short Solution Steps

Express 2\times \frac{1}{4x^{2}-8x} as a single fraction.

$$\frac{1}{x^{2}-3x}+\frac{2}{4x^{2}-8x}$$

Factor x^{2}-3x. Factor 4x^{2}-8x.

$$\frac{1}{x\left(x-3\right)}+\frac{2}{4x\left(x-2\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x-3\right) and 4x\left(x-2\right) is 4x\left(x-3\right)\left(x-2\right). Multiply \frac{1}{x\left(x-3\right)} times \frac{4\left(x-2\right)}{4\left(x-2\right)}. Multiply \frac{2}{4x\left(x-2\right)} times \frac{x-3}{x-3}.

$$\frac{4\left(x-2\right)}{4x\left(x-3\right)\left(x-2\right)}+\frac{2\left(x-3\right)}{4x\left(x-3\right)\left(x-2\right)}$$

Since \frac{4\left(x-2\right)}{4x\left(x-3\right)\left(x-2\right)} and \frac{2\left(x-3\right)}{4x\left(x-3\right)\left(x-2\right)} have the same denominator, add them by adding their numerators.

$$\frac{4\left(x-2\right)+2\left(x-3\right)}{4x\left(x-3\right)\left(x-2\right)}$$

Do the multiplications in 4\left(x-2\right)+2\left(x-3\right).

$$\frac{4x-8+2x-6}{4x\left(x-3\right)\left(x-2\right)}$$

Combine like terms in 4x-8+2x-6.

$$\frac{6x-14}{4x\left(x-3\right)\left(x-2\right)}$$

Factor the expressions that are not already factored in \frac{6x-14}{4x\left(x-3\right)\left(x-2\right)}.

$$\frac{2\left(3x-7\right)}{4x\left(x-3\right)\left(x-2\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{3x-7}{2x\left(x-3\right)\left(x-2\right)}$$

Expand 2x\left(x-3\right)\left(x-2\right).

$$\frac{3x-7}{2x^{3}-10x^{2}+12x}$$

Factor

$$\frac{3x-7}{2x\left(x-3\right)\left(x-2\right)}$$