Solve z^2-1/z^2-4-z-1/2z+4

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

Evaluate

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

Short Solution Steps

Factor z^{2}-4. Factor 2z+4.

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

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

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

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

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

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

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

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

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

Expand 2\left(z-2\right)\left(z+2\right).

$$\frac{z^{2}-4+3z}{2z^{2}-8}$$

Expand

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

Short Solution Steps

Factor z^{2}-4. Factor 2z+4.

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

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

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

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

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

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

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

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

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

Expand 2\left(z-2\right)\left(z+2\right).

$$\frac{z^{2}-4+3z}{2z^{2}-8}$$