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

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

Evaluate

$$-\frac{14x}{\left(x+2\right)\left(x-2\right)^{2}}$$

Short Solution Steps

Factor x^{2}-4.

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

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

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

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

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

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

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

Combine like terms in 2x-4+x+3.

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

Factor x^{2}-4x+4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)\left(x+2\right) and \left(x-2\right)^{2} is \left(x+2\right)\left(x-2\right)^{2}. Multiply \frac{3x-1}{\left(x-2\right)\left(x+2\right)} times \frac{x-2}{x-2}. Multiply \frac{3x+1}{\left(x-2\right)^{2}} times \frac{x+2}{x+2}.

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

Since \frac{\left(3x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^{2}} and \frac{\left(3x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(3x-1\right)\left(x-2\right)-\left(3x+1\right)\left(x+2\right).

$$\frac{3x^{2}-6x-x+2-3x^{2}-6x-x-2}{\left(x+2\right)\left(x-2\right)^{2}}$$

Combine like terms in 3x^{2}-6x-x+2-3x^{2}-6x-x-2.

$$\frac{-14x}{\left(x+2\right)\left(x-2\right)^{2}}$$

Expand \left(x+2\right)\left(x-2\right)^{2}.

$$\frac{-14x}{x^{3}-2x^{2}-4x+8}$$

Expand

$$-\frac{14x}{\left(x+2\right)\left(x-2\right)^{2}}$$

Short Solution Steps

Factor x^{2}-4.

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

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

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

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

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

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

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

Combine like terms in 2x-4+x+3.

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

Factor x^{2}-4x+4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)\left(x+2\right) and \left(x-2\right)^{2} is \left(x+2\right)\left(x-2\right)^{2}. Multiply \frac{3x-1}{\left(x-2\right)\left(x+2\right)} times \frac{x-2}{x-2}. Multiply \frac{3x+1}{\left(x-2\right)^{2}} times \frac{x+2}{x+2}.

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

Since \frac{\left(3x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^{2}} and \frac{\left(3x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(3x-1\right)\left(x-2\right)-\left(3x+1\right)\left(x+2\right).

$$\frac{3x^{2}-6x-x+2-3x^{2}-6x-x-2}{\left(x+2\right)\left(x-2\right)^{2}}$$

Combine like terms in 3x^{2}-6x-x+2-3x^{2}-6x-x-2.

$$\frac{-14x}{\left(x+2\right)\left(x-2\right)^{2}}$$

Expand \left(x+2\right)\left(x-2\right)^{2}.

$$\frac{-14x}{x^{3}-2x^{2}-4x+8}$$