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

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

Evaluate

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

Short Solution Steps

Factor 2a-4b. Factor 6a+4b.

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

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

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

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

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

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

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

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

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

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

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

Cancel out 2 in both numerator and denominator.

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

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

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

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

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

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

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

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

$$\frac{-3a^{2}+2ab-6ba+4b^{2}-3a^{2}+6ab}{\left(a-2b\right)\left(-3a+2b\right)\left(3a+2b\right)}$$

Combine like terms in -3a^{2}+2ab-6ba+4b^{2}-3a^{2}+6ab.

$$\frac{-6a^{2}+4b^{2}+2ab}{\left(a-2b\right)\left(-3a+2b\right)\left(3a+2b\right)}$$

Factor the expressions that are not already factored in \frac{-6a^{2}+4b^{2}+2ab}{\left(a-2b\right)\left(-3a+2b\right)\left(3a+2b\right)}.

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

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

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

Expand \left(a-2b\right)\left(-3a+2b\right).

$$\frac{2\left(-a+b\right)}{-3a^{2}+8ab-4b^{2}}$$

Use the distributive property to multiply 2 by -a+b.

$$\frac{-2a+2b}{-3a^{2}+8ab-4b^{2}}$$

Factor

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