Solve 1/3y-2-1/3y+1

$$\frac{ 1 }{ 3y-2 } - \frac{ 1 }{ 3y+1 }$$

Evaluate

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

Short Solution Steps

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

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

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

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

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

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

Combine like terms in 3y+1-3y+2.

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

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

$$\frac{3}{9y^{2}-3y-2}$$

Differentiate w.r.t. y

$$\frac{9\left(1-6y\right)}{\left(\left(3y-2\right)\left(3y+1\right)\right)^{2}}$$