Solve 4x/x-8-3+3/x

$$\frac{ 4x }{ x-8 } -3+ \frac{ 3 }{ x }$$

Evaluate

$$\frac{x^{2}+27x-24}{x\left(x-8\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-8}{x-8}.

$$\frac{4x}{x-8}-\frac{3\left(x-8\right)}{x-8}+\frac{3}{x}$$

Since \frac{4x}{x-8} and \frac{3\left(x-8\right)}{x-8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4x-3\left(x-8\right)}{x-8}+\frac{3}{x}$$

Do the multiplications in 4x-3\left(x-8\right).

$$\frac{4x-3x+24}{x-8}+\frac{3}{x}$$

Combine like terms in 4x-3x+24.

$$\frac{x+24}{x-8}+\frac{3}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-8 and x is x\left(x-8\right). Multiply \frac{x+24}{x-8} times \frac{x}{x}. Multiply \frac{3}{x} times \frac{x-8}{x-8}.

$$\frac{\left(x+24\right)x}{x\left(x-8\right)}+\frac{3\left(x-8\right)}{x\left(x-8\right)}$$

Since \frac{\left(x+24\right)x}{x\left(x-8\right)} and \frac{3\left(x-8\right)}{x\left(x-8\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+24\right)x+3\left(x-8\right)}{x\left(x-8\right)}$$

Do the multiplications in \left(x+24\right)x+3\left(x-8\right).

$$\frac{x^{2}+24x+3x-24}{x\left(x-8\right)}$$

Combine like terms in x^{2}+24x+3x-24.

$$\frac{x^{2}+27x-24}{x\left(x-8\right)}$$

Expand x\left(x-8\right).

$$\frac{x^{2}+27x-24}{x^{2}-8x}$$

Differentiate w.r.t. x

$$\frac{-35x^{2}+48x-192}{\left(x\left(x-8\right)\right)^{2}}$$