Solve t+1/t^2+5t+8/t^2-254

$$\frac { t + 1 } { t ^ { 2 } + 5 t } + \frac { 8 } { t ^ { 2 } - 25 } 4$$

Evaluate

$$\frac{t^{2}+28t-5}{t\left(t^{2}-25\right)}$$

Short Solution Steps

Express \frac{8}{t^{2}-25}\times 4 as a single fraction.

$$\frac{t+1}{t^{2}+5t}+\frac{8\times 4}{t^{2}-25}$$

Factor t^{2}+5t. Factor t^{2}-25.

$$\frac{t+1}{t\left(t+5\right)}+\frac{8\times 4}{\left(t-5\right)\left(t+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t\left(t+5\right) and \left(t-5\right)\left(t+5\right) is t\left(t-5\right)\left(t+5\right). Multiply \frac{t+1}{t\left(t+5\right)} times \frac{t-5}{t-5}. Multiply \frac{8\times 4}{\left(t-5\right)\left(t+5\right)} times \frac{t}{t}.

$$\frac{\left(t+1\right)\left(t-5\right)}{t\left(t-5\right)\left(t+5\right)}+\frac{8\times 4t}{t\left(t-5\right)\left(t+5\right)}$$

Since \frac{\left(t+1\right)\left(t-5\right)}{t\left(t-5\right)\left(t+5\right)} and \frac{8\times 4t}{t\left(t-5\right)\left(t+5\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(t+1\right)\left(t-5\right)+8\times 4t}{t\left(t-5\right)\left(t+5\right)}$$

Do the multiplications in \left(t+1\right)\left(t-5\right)+8\times 4t.

$$\frac{t^{2}-5t+t-5+32t}{t\left(t-5\right)\left(t+5\right)}$$

Combine like terms in t^{2}-5t+t-5+32t.

$$\frac{t^{2}+28t-5}{t\left(t-5\right)\left(t+5\right)}$$

Expand t\left(t-5\right)\left(t+5\right).

$$\frac{t^{2}+28t-5}{t^{3}-25t}$$

Expand

$$\frac{t^{2}+28t-5}{t\left(t^{2}-25\right)}$$

Short Solution Steps

Express \frac{8}{t^{2}-25}\times 4 as a single fraction.

$$\frac{t+1}{t^{2}+5t}+\frac{8\times 4}{t^{2}-25}$$

Factor t^{2}+5t. Factor t^{2}-25.

$$\frac{t+1}{t\left(t+5\right)}+\frac{8\times 4}{\left(t-5\right)\left(t+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t\left(t+5\right) and \left(t-5\right)\left(t+5\right) is t\left(t-5\right)\left(t+5\right). Multiply \frac{t+1}{t\left(t+5\right)} times \frac{t-5}{t-5}. Multiply \frac{8\times 4}{\left(t-5\right)\left(t+5\right)} times \frac{t}{t}.

$$\frac{\left(t+1\right)\left(t-5\right)}{t\left(t-5\right)\left(t+5\right)}+\frac{8\times 4t}{t\left(t-5\right)\left(t+5\right)}$$

Since \frac{\left(t+1\right)\left(t-5\right)}{t\left(t-5\right)\left(t+5\right)} and \frac{8\times 4t}{t\left(t-5\right)\left(t+5\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(t+1\right)\left(t-5\right)+8\times 4t}{t\left(t-5\right)\left(t+5\right)}$$

Do the multiplications in \left(t+1\right)\left(t-5\right)+8\times 4t.

$$\frac{t^{2}-5t+t-5+32t}{t\left(t-5\right)\left(t+5\right)}$$

Combine like terms in t^{2}-5t+t-5+32t.

$$\frac{t^{2}+28t-5}{t\left(t-5\right)\left(t+5\right)}$$

Expand t\left(t-5\right)\left(t+5\right).

$$\frac{t^{2}+28t-5}{t^{3}-25t}$$