Solve x/x+3divx^2+x/x^2+bx+8+3x-3/x^2-1

$$\frac { x } { x + 3 } \div \frac { x ^ { 2 } + x } { x ^ { 2 } + b x + 8 } + \frac { 3 x - 3 } { x ^ { 2 } - 1 }$$

Evaluate

$$\frac{x^{2}+bx+3x+17}{\left(x+1\right)\left(x+3\right)}$$

Short Solution Steps

Divide \frac{x}{x+3} by \frac{x^{2}+x}{x^{2}+bx+8} by multiplying \frac{x}{x+3} by the reciprocal of \frac{x^{2}+x}{x^{2}+bx+8}.

$$\frac{x\left(x^{2}+bx+8\right)}{\left(x+3\right)\left(x^{2}+x\right)}+\frac{3x-3}{x^{2}-1}$$

Factor the expressions that are not already factored in \frac{x\left(x^{2}+bx+8\right)}{\left(x+3\right)\left(x^{2}+x\right)}.

$$\frac{x\left(x^{2}+bx+8\right)}{x\left(x+1\right)\left(x+3\right)}+\frac{3x-3}{x^{2}-1}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3x-3}{x^{2}-1}$$

Factor the expressions that are not already factored in \frac{3x-3}{x^{2}-1}.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}$$

Cancel out x-1 in both numerator and denominator.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3}{x+1}$$

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

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}$$

Since \frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)} and \frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+bx+8+3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}$$

Do the multiplications in x^{2}+bx+8+3\left(x+3\right).

$$\frac{x^{2}+bx+8+3x+9}{\left(x+1\right)\left(x+3\right)}$$

Combine like terms in x^{2}+bx+8+3x+9.

$$\frac{x^{2}+bx+17+3x}{\left(x+1\right)\left(x+3\right)}$$

Expand \left(x+1\right)\left(x+3\right).

$$\frac{x^{2}+bx+17+3x}{x^{2}+4x+3}$$

Expand

$$\frac{x^{2}+bx+3x+17}{\left(x+1\right)\left(x+3\right)}$$

Short Solution Steps

Divide \frac{x}{x+3} by \frac{x^{2}+x}{x^{2}+bx+8} by multiplying \frac{x}{x+3} by the reciprocal of \frac{x^{2}+x}{x^{2}+bx+8}.

$$\frac{x\left(x^{2}+bx+8\right)}{\left(x+3\right)\left(x^{2}+x\right)}+\frac{3x-3}{x^{2}-1}$$

Factor the expressions that are not already factored in \frac{x\left(x^{2}+bx+8\right)}{\left(x+3\right)\left(x^{2}+x\right)}.

$$\frac{x\left(x^{2}+bx+8\right)}{x\left(x+1\right)\left(x+3\right)}+\frac{3x-3}{x^{2}-1}$$

Cancel out x in both numerator and denominator.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3x-3}{x^{2}-1}$$

Factor the expressions that are not already factored in \frac{3x-3}{x^{2}-1}.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}$$

Cancel out x-1 in both numerator and denominator.

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3}{x+1}$$

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

$$\frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}$$

Since \frac{x^{2}+bx+8}{\left(x+1\right)\left(x+3\right)} and \frac{3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+bx+8+3\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}$$

Do the multiplications in x^{2}+bx+8+3\left(x+3\right).

$$\frac{x^{2}+bx+8+3x+9}{\left(x+1\right)\left(x+3\right)}$$

Combine like terms in x^{2}+bx+8+3x+9.

$$\frac{x^{2}+bx+17+3x}{\left(x+1\right)\left(x+3\right)}$$

Expand \left(x+1\right)\left(x+3\right).

$$\frac{x^{2}+bx+17+3x}{x^{2}+4x+3}$$