Solve 3/3x-5+2/3-2x

$$\frac{ 3 }{ 3x-5 } + \frac{ 2 }{ 3-2x }$$

Evaluate

$$-\frac{1}{\left(3-2x\right)\left(3x-5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x-5 and 3-2x is \left(3x-5\right)\left(-2x+3\right). Multiply \frac{3}{3x-5} times \frac{-2x+3}{-2x+3}. Multiply \frac{2}{3-2x} times \frac{3x-5}{3x-5}.

$$\frac{3\left(-2x+3\right)}{\left(3x-5\right)\left(-2x+3\right)}+\frac{2\left(3x-5\right)}{\left(3x-5\right)\left(-2x+3\right)}$$

Since \frac{3\left(-2x+3\right)}{\left(3x-5\right)\left(-2x+3\right)} and \frac{2\left(3x-5\right)}{\left(3x-5\right)\left(-2x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\left(-2x+3\right)+2\left(3x-5\right)}{\left(3x-5\right)\left(-2x+3\right)}$$

Do the multiplications in 3\left(-2x+3\right)+2\left(3x-5\right).

$$\frac{-6x+9+6x-10}{\left(3x-5\right)\left(-2x+3\right)}$$

Combine like terms in -6x+9+6x-10.

$$\frac{-1}{\left(3x-5\right)\left(-2x+3\right)}$$

Expand \left(3x-5\right)\left(-2x+3\right).

$$\frac{-1}{-6x^{2}+19x-15}$$

Differentiate w.r.t. x

$$\frac{12x-19}{-36x^{4}+228x^{3}-541x^{2}+570x-225}$$