Solve 2x+4/6x^2+3x-(2(x+2)/3x(x+1))

$$\frac { 2 x + 4 } { 6 x ^ { 2 } + 3 x } - ( \frac { 2 ( x + 2 ) } { 3 x ( x + 1 ) } )$$

Evaluate

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

Short Solution Steps

Factor 6x^{2}+3x.

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

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

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

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

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

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

$$\frac{2x^{2}+2x+4x+4-4x^{2}-2x-8x-4}{3x\left(x+1\right)\left(2x+1\right)}$$

Combine like terms in 2x^{2}+2x+4x+4-4x^{2}-2x-8x-4.

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

Factor the expressions that are not already factored in \frac{-2x^{2}-4x}{3x\left(x+1\right)\left(2x+1\right)}.

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

Cancel out x in both numerator and denominator.

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

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

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

Use the distributive property to multiply 2 by -x-2.

$$\frac{-2x-4}{6x^{2}+9x+3}$$

Expand

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

Short Solution Steps

Factor 6x^{2}+3x.

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

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

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

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

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

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

$$\frac{2x^{2}+2x+4x+4-4x^{2}-2x-8x-4}{3x\left(x+1\right)\left(2x+1\right)}$$

Combine like terms in 2x^{2}+2x+4x+4-4x^{2}-2x-8x-4.

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

Factor the expressions that are not already factored in \frac{-2x^{2}-4x}{3x\left(x+1\right)\left(2x+1\right)}.

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

Cancel out x in both numerator and denominator.

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

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

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

Use the distributive property to multiply 2 by -x-2.

$$\frac{-2x-4}{6x^{2}+9x+3}$$