Solve 4/a-3+3/a+3-24/a^2-9

$$\frac{ 4 }{ a-3 } + \frac{ 3 }{ a+3 } - \frac{ 24 }{ { a }^{ 2 } -9 }$$

Evaluate

$$\frac{7}{a+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a-3 and a+3 is \left(a-3\right)\left(a+3\right). Multiply \frac{4}{a-3} times \frac{a+3}{a+3}. Multiply \frac{3}{a+3} times \frac{a-3}{a-3}.

$$\frac{4\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}+\frac{3\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}-\frac{24}{a^{2}-9}$$

Since \frac{4\left(a+3\right)}{\left(a-3\right)\left(a+3\right)} and \frac{3\left(a-3\right)}{\left(a-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{4\left(a+3\right)+3\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}-\frac{24}{a^{2}-9}$$

Do the multiplications in 4\left(a+3\right)+3\left(a-3\right).

$$\frac{4a+12+3a-9}{\left(a-3\right)\left(a+3\right)}-\frac{24}{a^{2}-9}$$

Combine like terms in 4a+12+3a-9.

$$\frac{7a+3}{\left(a-3\right)\left(a+3\right)}-\frac{24}{a^{2}-9}$$

Factor a^{2}-9.

$$\frac{7a+3}{\left(a-3\right)\left(a+3\right)}-\frac{24}{\left(a-3\right)\left(a+3\right)}$$

Since \frac{7a+3}{\left(a-3\right)\left(a+3\right)} and \frac{24}{\left(a-3\right)\left(a+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7a+3-24}{\left(a-3\right)\left(a+3\right)}$$

Combine like terms in 7a+3-24.

$$\frac{7a-21}{\left(a-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored in \frac{7a-21}{\left(a-3\right)\left(a+3\right)}.

$$\frac{7\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{7}{a+3}$$

Differentiate w.r.t. a

$$-\frac{7}{\left(a+3\right)^{2}}$$