Solve 2z+3/z^2+4z-12+7/z^2+5z-6

$$\frac { 2 z + 3 } { z ^ { 2 } + 4 z - 12 } + \frac { 7 } { z ^ { 2 } + 5 z - 6 }$$

Evaluate

$$\frac{2z^{2}+8z-17}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Short Solution Steps

Factor z^{2}+4z-12. Factor z^{2}+5z-6.

$$\frac{2z+3}{\left(z-2\right)\left(z+6\right)}+\frac{7}{\left(z-1\right)\left(z+6\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(z-2\right)\left(z+6\right) and \left(z-1\right)\left(z+6\right) is \left(z-2\right)\left(z-1\right)\left(z+6\right). Multiply \frac{2z+3}{\left(z-2\right)\left(z+6\right)} times \frac{z-1}{z-1}. Multiply \frac{7}{\left(z-1\right)\left(z+6\right)} times \frac{z-2}{z-2}.

$$\frac{\left(2z+3\right)\left(z-1\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}+\frac{7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Since \frac{\left(2z+3\right)\left(z-1\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)} and \frac{7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2z+3\right)\left(z-1\right)+7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Do the multiplications in \left(2z+3\right)\left(z-1\right)+7\left(z-2\right).

$$\frac{2z^{2}-2z+3z-3+7z-14}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Combine like terms in 2z^{2}-2z+3z-3+7z-14.

$$\frac{2z^{2}+8z-17}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Expand \left(z-2\right)\left(z-1\right)\left(z+6\right).

$$\frac{2z^{2}+8z-17}{z^{3}+3z^{2}-16z+12}$$

Expand

$$\frac{2z^{2}+8z-17}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Short Solution Steps

Factor z^{2}+4z-12. Factor z^{2}+5z-6.

$$\frac{2z+3}{\left(z-2\right)\left(z+6\right)}+\frac{7}{\left(z-1\right)\left(z+6\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(z-2\right)\left(z+6\right) and \left(z-1\right)\left(z+6\right) is \left(z-2\right)\left(z-1\right)\left(z+6\right). Multiply \frac{2z+3}{\left(z-2\right)\left(z+6\right)} times \frac{z-1}{z-1}. Multiply \frac{7}{\left(z-1\right)\left(z+6\right)} times \frac{z-2}{z-2}.

$$\frac{\left(2z+3\right)\left(z-1\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}+\frac{7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Since \frac{\left(2z+3\right)\left(z-1\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)} and \frac{7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2z+3\right)\left(z-1\right)+7\left(z-2\right)}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Do the multiplications in \left(2z+3\right)\left(z-1\right)+7\left(z-2\right).

$$\frac{2z^{2}-2z+3z-3+7z-14}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Combine like terms in 2z^{2}-2z+3z-3+7z-14.

$$\frac{2z^{2}+8z-17}{\left(z-2\right)\left(z-1\right)\left(z+6\right)}$$

Expand \left(z-2\right)\left(z-1\right)\left(z+6\right).

$$\frac{2z^{2}+8z-17}{z^{3}+3z^{2}-16z+12}$$