Solve 6z^2/3z-1+2z^2/3z+1

$$\frac { 6 z ^ { 2 } } { 3 z - 1 } + \frac { 2 z ^ { 2 } } { 3 z + 1 }$$

Evaluate

$$\frac{4\left(6z+1\right)z^{2}}{9z^{2}-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3z-1 and 3z+1 is \left(3z-1\right)\left(3z+1\right). Multiply \frac{6z^{2}}{3z-1} times \frac{3z+1}{3z+1}. Multiply \frac{2z^{2}}{3z+1} times \frac{3z-1}{3z-1}.

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

Since \frac{6z^{2}\left(3z+1\right)}{\left(3z-1\right)\left(3z+1\right)} and \frac{2z^{2}\left(3z-1\right)}{\left(3z-1\right)\left(3z+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 6z^{2}\left(3z+1\right)+2z^{2}\left(3z-1\right).

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

Combine like terms in 18z^{3}+6z^{2}+6z^{3}-2z^{2}.

$$\frac{24z^{3}+4z^{2}}{\left(3z-1\right)\left(3z+1\right)}$$

Expand \left(3z-1\right)\left(3z+1\right).

$$\frac{24z^{3}+4z^{2}}{9z^{2}-1}$$

Factor

$$\frac{4\left(6z+1\right)z^{2}}{9z^{2}-1}$$