Solve 1/6a-4b-1/6a+4b-3a/4b^2-9a^2

$$\frac { 1 } { 6 a - 4 b } - \frac { 1 } { 6 a + 4 b } - \frac { 3 a } { 4 b ^ { 2 } - 9 a ^ { 2 } }$$

Evaluate

$$\frac{1}{3a-2b}$$

Short Solution Steps

Factor 6a-4b. Factor 6a+4b.

$$\frac{1}{2\left(3a-2b\right)}-\frac{1}{2\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

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

$$\frac{3a+2b}{2\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a-2b}{2\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

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

$$\frac{3a+2b-\left(3a-2b\right)}{2\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

Do the multiplications in 3a+2b-\left(3a-2b\right).

$$\frac{3a+2b-3a+2b}{2\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

Combine like terms in 3a+2b-3a+2b.

$$\frac{4b}{2\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

Cancel out 2 in both numerator and denominator.

$$\frac{2b}{\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{4b^{2}-9a^{2}}$$

Factor 4b^{2}-9a^{2}.

$$\frac{2b}{\left(3a-2b\right)\left(3a+2b\right)}-\frac{3a}{\left(-3a+2b\right)\left(3a+2b\right)}$$

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

$$\frac{-2b}{\left(-3a+2b\right)\left(3a+2b\right)}-\frac{3a}{\left(-3a+2b\right)\left(3a+2b\right)}$$

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

$$\frac{-2b-3a}{\left(-3a+2b\right)\left(3a+2b\right)}$$

Extract the negative sign in -2b-3a.

$$\frac{-\left(3a+2b\right)}{\left(-3a+2b\right)\left(3a+2b\right)}$$

Cancel out 3a+2b in both numerator and denominator.

$$\frac{-1}{-3a+2b}$$

Factor

$$\frac{1}{3a-2b}$$