Solve a^2-9/2a^2+1*(6^2+1/a-3+6a-1/a+3)

$$\frac { a ^ { 2 } - 9 } { 2 a ^ { 2 } + 1 } \cdot ( \frac { 6 ^ { 2 } + 1 } { a - 3 } + \frac { 6 a - 1 } { a + 3 } )$$

Evaluate

$$\frac{6\left(a^{2}+3a+19\right)}{2a^{2}+1}$$

Short Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{36+1}{a-3}+\frac{6a-1}{a+3}\right)$$

Add 36 and 1 to get 37.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{37}{a-3}+\frac{6a-1}{a+3}\right)$$

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{37}{a-3} times \frac{a+3}{a+3}. Multiply \frac{6a-1}{a+3} times \frac{a-3}{a-3}.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{37\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}+\frac{\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)$$

Since \frac{37\left(a+3\right)}{\left(a-3\right)\left(a+3\right)} and \frac{\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{37\left(a+3\right)+\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}$$

Do the multiplications in 37\left(a+3\right)+\left(6a-1\right)\left(a-3\right).

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{37a+111+6a^{2}-18a-a+3}{\left(a-3\right)\left(a+3\right)}$$

Combine like terms in 37a+111+6a^{2}-18a-a+3.

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{18a+114+6a^{2}}{\left(a-3\right)\left(a+3\right)}$$

Multiply \frac{a^{2}-9}{2a^{2}+1} times \frac{18a+114+6a^{2}}{\left(a-3\right)\left(a+3\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-9\right)\left(18a+114+6a^{2}\right)}{\left(2a^{2}+1\right)\left(a-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{6\left(a-3\right)\left(a+3\right)\left(a^{2}+3a+19\right)}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Cancel out \left(a-3\right)\left(a+3\right) in both numerator and denominator.

$$\frac{6\left(a^{2}+3a+19\right)}{2a^{2}+1}$$

Expand the expression.

$$\frac{6a^{2}+18a+114}{2a^{2}+1}$$

Expand

$$\frac{6\left(a^{2}+3a+19\right)}{2a^{2}+1}$$

Short Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{36+1}{a-3}+\frac{6a-1}{a+3}\right)$$

Add 36 and 1 to get 37.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{37}{a-3}+\frac{6a-1}{a+3}\right)$$

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{37}{a-3} times \frac{a+3}{a+3}. Multiply \frac{6a-1}{a+3} times \frac{a-3}{a-3}.

$$\frac{a^{2}-9}{2a^{2}+1}\left(\frac{37\left(a+3\right)}{\left(a-3\right)\left(a+3\right)}+\frac{\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)$$

Since \frac{37\left(a+3\right)}{\left(a-3\right)\left(a+3\right)} and \frac{\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{37\left(a+3\right)+\left(6a-1\right)\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}$$

Do the multiplications in 37\left(a+3\right)+\left(6a-1\right)\left(a-3\right).

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{37a+111+6a^{2}-18a-a+3}{\left(a-3\right)\left(a+3\right)}$$

Combine like terms in 37a+111+6a^{2}-18a-a+3.

$$\frac{a^{2}-9}{2a^{2}+1}\times \frac{18a+114+6a^{2}}{\left(a-3\right)\left(a+3\right)}$$

Multiply \frac{a^{2}-9}{2a^{2}+1} times \frac{18a+114+6a^{2}}{\left(a-3\right)\left(a+3\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-9\right)\left(18a+114+6a^{2}\right)}{\left(2a^{2}+1\right)\left(a-3\right)\left(a+3\right)}$$

Factor the expressions that are not already factored.

$$\frac{6\left(a-3\right)\left(a+3\right)\left(a^{2}+3a+19\right)}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Cancel out \left(a-3\right)\left(a+3\right) in both numerator and denominator.

$$\frac{6\left(a^{2}+3a+19\right)}{2a^{2}+1}$$

Expand the expression.

$$\frac{6a^{2}+18a+114}{2a^{2}+1}$$