Solve 2y-8/y^2-16-y/y-1+y^2+2/y^2+3y-4

$$\frac { 2 y - 8 } { y ^ { 2 } - 16 } - \frac { y } { y - 1 } + \frac { y ^ { 2 } + 2 } { y ^ { 2 } + 3 y - 4 }$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{2y-8}{y^{2}-16}.

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

Cancel out y-4 in both numerator and denominator.

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

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

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

Since \frac{2\left(y-1\right)}{\left(y-1\right)\left(y+4\right)} and \frac{y\left(y+4\right)}{\left(y-1\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(y-1\right)-y\left(y+4\right).

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

Combine like terms in 2y-2-y^{2}-4y.

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

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

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

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

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

Combine like terms in -2y-2-y^{2}+y^{2}+2.

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

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

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

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{2y-8}{y^{2}-16}.

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

Cancel out y-4 in both numerator and denominator.

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

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

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

Since \frac{2\left(y-1\right)}{\left(y-1\right)\left(y+4\right)} and \frac{y\left(y+4\right)}{\left(y-1\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(y-1\right)-y\left(y+4\right).

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

Combine like terms in 2y-2-y^{2}-4y.

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

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

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

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

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

Combine like terms in -2y-2-y^{2}+y^{2}+2.

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

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

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