Solve 3y/y+4-y+1/y^2-16=

$$\frac { 3 y } { y + 4 } - \frac { y + 1 } { y ^ { 2 } - 16 } =$$

Evaluate

$$\frac{3y^{2}-13y-1}{y^{2}-16}$$

Short Solution Steps

Factor y^{2}-16.

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

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

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

Since \frac{3y\left(y-4\right)}{\left(y-4\right)\left(y+4\right)} and \frac{y+1}{\left(y-4\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{3y^{2}-12y-y-1}{\left(y-4\right)\left(y+4\right)}$$

Combine like terms in 3y^{2}-12y-y-1.

$$\frac{3y^{2}-13y-1}{\left(y-4\right)\left(y+4\right)}$$

Expand \left(y-4\right)\left(y+4\right).

$$\frac{3y^{2}-13y-1}{y^{2}-16}$$

Expand

$$\frac{3y^{2}-13y-1}{y^{2}-16}$$

Short Solution Steps

Factor y^{2}-16.

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

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

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

Since \frac{3y\left(y-4\right)}{\left(y-4\right)\left(y+4\right)} and \frac{y+1}{\left(y-4\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{3y^{2}-12y-y-1}{\left(y-4\right)\left(y+4\right)}$$

Combine like terms in 3y^{2}-12y-y-1.

$$\frac{3y^{2}-13y-1}{\left(y-4\right)\left(y+4\right)}$$

Expand \left(y-4\right)\left(y+4\right).

$$\frac{3y^{2}-13y-1}{y^{2}-16}$$