Solve 2a/z-3-2a/a+3+36/a^2-9

$$\frac { 2 a } { z - 3 } - \frac { 2 a } { a + 3 } + \frac { 36 } { a ^ { 2 } - 9 }$$

Evaluate

$$\frac{2\left(-az+6z+a^{2}-18\right)}{\left(a-3\right)\left(z-3\right)}$$

Short Solution Steps

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

$$\frac{2a\left(a+3\right)}{\left(z-3\right)\left(a+3\right)}-\frac{2a\left(z-3\right)}{\left(z-3\right)\left(a+3\right)}+\frac{36}{a^{2}-9}$$

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

$$\frac{2a\left(a+3\right)-2a\left(z-3\right)}{\left(z-3\right)\left(a+3\right)}+\frac{36}{a^{2}-9}$$

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

$$\frac{2a^{2}+6a-2az+6a}{\left(z-3\right)\left(a+3\right)}+\frac{36}{a^{2}-9}$$

Combine like terms in 2a^{2}+6a-2az+6a.

$$\frac{2a^{2}+12a-2az}{\left(z-3\right)\left(a+3\right)}+\frac{36}{a^{2}-9}$$

Factor a^{2}-9.

$$\frac{2a^{2}+12a-2az}{\left(z-3\right)\left(a+3\right)}+\frac{36}{\left(a-3\right)\left(a+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(z-3\right)\left(a+3\right) and \left(a-3\right)\left(a+3\right) is \left(a-3\right)\left(z-3\right)\left(a+3\right). Multiply \frac{2a^{2}+12a-2az}{\left(z-3\right)\left(a+3\right)} times \frac{a-3}{a-3}. Multiply \frac{36}{\left(a-3\right)\left(a+3\right)} times \frac{z-3}{z-3}.

$$\frac{\left(2a^{2}+12a-2az\right)\left(a-3\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}+\frac{36\left(z-3\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}$$

Since \frac{\left(2a^{2}+12a-2az\right)\left(a-3\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)} and \frac{36\left(z-3\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2a^{2}+12a-2az\right)\left(a-3\right)+36\left(z-3\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}$$

Do the multiplications in \left(2a^{2}+12a-2az\right)\left(a-3\right)+36\left(z-3\right).

$$\frac{2a^{3}-6a^{2}+12a^{2}-36a-2a^{2}z+6az+36z-108}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}$$

Combine like terms in 2a^{3}-6a^{2}+12a^{2}-36a-2a^{2}z+6az+36z-108.

$$\frac{2a^{3}+6a^{2}-36a+36z-2a^{2}z+6az-108}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored in \frac{2a^{3}+6a^{2}-36a+36z-2a^{2}z+6az-108}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}.

$$\frac{2\left(a+3\right)\left(-az+6z+a^{2}-18\right)}{\left(a-3\right)\left(z-3\right)\left(a+3\right)}$$

Cancel out a+3 in both numerator and denominator.

$$\frac{2\left(-az+6z+a^{2}-18\right)}{\left(a-3\right)\left(z-3\right)}$$

Expand \left(a-3\right)\left(z-3\right).

$$\frac{2\left(-az+6z+a^{2}-18\right)}{az-3z-3a+9}$$

Use the distributive property to multiply 2 by -az+6z+a^{2}-18.

$$\frac{-2az+12z+2a^{2}-36}{az-3z-3a+9}$$

Factor

$$\frac{2\left(-az+6z+a^{2}-18\right)}{\left(a-3\right)\left(z-3\right)}$$