Solve 2x+1/12x+8-x^2/6x^2+x-2+2x/16x-8

$$\frac { 2 x + 1 } { 12 x + 8 } - \frac { x ^ { 2 } } { 6 x ^ { 2 } + x - 2 } + \frac { 2 x } { 16 x - 8 }$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{2x}{16x-8}.

$$\frac{2x+1}{12x+8}-\frac{x^{2}}{6x^{2}+x-2}+\frac{2x}{8\left(2x-1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{2x+1}{12x+8}-\frac{x^{2}}{6x^{2}+x-2}+\frac{x}{4\left(2x-1\right)}$$

Factor 12x+8. Factor 6x^{2}+x-2.

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

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

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

Since \frac{\left(2x+1\right)\left(2x-1\right)}{4\left(2x-1\right)\left(3x+2\right)} and \frac{4x^{2}}{4\left(2x-1\right)\left(3x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(2x+1\right)\left(2x-1\right)-4x^{2}.

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

Combine like terms in 4x^{2}-2x+2x-1-4x^{2}.

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

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

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

Since \frac{-1}{4\left(2x-1\right)\left(3x+2\right)} and \frac{x\left(3x+2\right)}{4\left(2x-1\right)\left(3x+2\right)} have the same denominator, add them by adding their numerators.

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

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

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

Expand 4\left(2x-1\right)\left(3x+2\right).

$$\frac{-1+3x^{2}+2x}{24x^{2}+4x-8}$$

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{2x}{16x-8}.

$$\frac{2x+1}{12x+8}-\frac{x^{2}}{6x^{2}+x-2}+\frac{2x}{8\left(2x-1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{2x+1}{12x+8}-\frac{x^{2}}{6x^{2}+x-2}+\frac{x}{4\left(2x-1\right)}$$

Factor 12x+8. Factor 6x^{2}+x-2.

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

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

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

Since \frac{\left(2x+1\right)\left(2x-1\right)}{4\left(2x-1\right)\left(3x+2\right)} and \frac{4x^{2}}{4\left(2x-1\right)\left(3x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(2x+1\right)\left(2x-1\right)-4x^{2}.

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

Combine like terms in 4x^{2}-2x+2x-1-4x^{2}.

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

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

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

Since \frac{-1}{4\left(2x-1\right)\left(3x+2\right)} and \frac{x\left(3x+2\right)}{4\left(2x-1\right)\left(3x+2\right)} have the same denominator, add them by adding their numerators.

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

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

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

Expand 4\left(2x-1\right)\left(3x+2\right).

$$\frac{-1+3x^{2}+2x}{24x^{2}+4x-8}$$