Solve =n-2/n-1+3/n^2+3n-4

$$= \frac { n - 2 } { n - 1 } + \frac { 3 } { n ^ { 2 } + 3 n - 4 }$$

Evaluate

$$\frac{n^{2}+2n-5}{\left(n-1\right)\left(n+4\right)}$$

Short Solution Steps

Factor n^{2}+3n-4.

$$\frac{n-2}{n-1}+\frac{3}{\left(n-1\right)\left(n+4\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-1 and \left(n-1\right)\left(n+4\right) is \left(n-1\right)\left(n+4\right). Multiply \frac{n-2}{n-1} times \frac{n+4}{n+4}.

$$\frac{\left(n-2\right)\left(n+4\right)}{\left(n-1\right)\left(n+4\right)}+\frac{3}{\left(n-1\right)\left(n+4\right)}$$

Since \frac{\left(n-2\right)\left(n+4\right)}{\left(n-1\right)\left(n+4\right)} and \frac{3}{\left(n-1\right)\left(n+4\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(n-2\right)\left(n+4\right)+3}{\left(n-1\right)\left(n+4\right)}$$

Do the multiplications in \left(n-2\right)\left(n+4\right)+3.

$$\frac{n^{2}+4n-2n-8+3}{\left(n-1\right)\left(n+4\right)}$$

Combine like terms in n^{2}+4n-2n-8+3.

$$\frac{n^{2}+2n-5}{\left(n-1\right)\left(n+4\right)}$$

Expand \left(n-1\right)\left(n+4\right).

$$\frac{n^{2}+2n-5}{n^{2}+3n-4}$$

Expand

$$\frac{n^{2}+2n-5}{\left(n-1\right)\left(n+4\right)}$$

Short Solution Steps

Factor n^{2}+3n-4.

$$\frac{n-2}{n-1}+\frac{3}{\left(n-1\right)\left(n+4\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-1 and \left(n-1\right)\left(n+4\right) is \left(n-1\right)\left(n+4\right). Multiply \frac{n-2}{n-1} times \frac{n+4}{n+4}.

$$\frac{\left(n-2\right)\left(n+4\right)}{\left(n-1\right)\left(n+4\right)}+\frac{3}{\left(n-1\right)\left(n+4\right)}$$

Since \frac{\left(n-2\right)\left(n+4\right)}{\left(n-1\right)\left(n+4\right)} and \frac{3}{\left(n-1\right)\left(n+4\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(n-2\right)\left(n+4\right)+3}{\left(n-1\right)\left(n+4\right)}$$

Do the multiplications in \left(n-2\right)\left(n+4\right)+3.

$$\frac{n^{2}+4n-2n-8+3}{\left(n-1\right)\left(n+4\right)}$$

Combine like terms in n^{2}+4n-2n-8+3.

$$\frac{n^{2}+2n-5}{\left(n-1\right)\left(n+4\right)}$$

Expand \left(n-1\right)\left(n+4\right).

$$\frac{n^{2}+2n-5}{n^{2}+3n-4}$$