Solve x/2x^2+7x+3-3/3x^2+7x-6

$$\frac{ x }{ 2 { x }^{ 2 } +7x+3 } - \frac{ 3 }{ 3 { x }^{ 2 } +7x-6 }$$

Evaluate

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

Short Solution Steps

Factor 2x^{2}+7x+3. Factor 3x^{2}+7x-6.

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

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

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

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

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

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

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

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

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

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

$$\frac{3x^{2}-8x-3}{6x^{3}+17x^{2}-5x-6}$$

Differentiate w.r.t. x

$$\frac{33+66x+175x^{2}+96x^{3}-18x^{4}}{\left(\left(3x-2\right)\left(x+3\right)\left(2x+1\right)\right)^{2}}$$