Solve 2/6+3n/n^2-5n+4

$$\frac { 2 } { 6 } + \frac { 3 n } { n ^ { 2 } - 5 n + 4 }$$

Evaluate

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

Short Solution Steps

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$\frac{1}{3}+\frac{3n}{n^{2}-5n+4}$$

Factor n^{2}-5n+4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and \left(n-4\right)\left(n-1\right) is 3\left(n-4\right)\left(n-1\right). Multiply \frac{1}{3} times \frac{\left(n-4\right)\left(n-1\right)}{\left(n-4\right)\left(n-1\right)}. Multiply \frac{3n}{\left(n-4\right)\left(n-1\right)} times \frac{3}{3}.

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

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

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

Do the multiplications in \left(n-4\right)\left(n-1\right)+3\times 3n.

$$\frac{n^{2}-n-4n+4+9n}{3\left(n-4\right)\left(n-1\right)}$$

Combine like terms in n^{2}-n-4n+4+9n.

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

Expand 3\left(n-4\right)\left(n-1\right).

$$\frac{n^{2}+4n+4}{3n^{2}-15n+12}$$

Factor

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