Solve 3/x-4+8/(x+4)(3-2x)

$$\frac { 3 } { x - 4 } + \frac { 8 } { ( x + 4 ) ( 3 - 2 x ) }$$

Evaluate

$$\frac{4-7x-6x^{2}}{\left(3-2x\right)\left(x^{2}-16\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-4 and \left(x+4\right)\left(3-2x\right) is \left(x-4\right)\left(x+4\right)\left(-2x+3\right). Multiply \frac{3}{x-4} times \frac{\left(x+4\right)\left(-2x+3\right)}{\left(x+4\right)\left(-2x+3\right)}. Multiply \frac{8}{\left(x+4\right)\left(3-2x\right)} times \frac{x-4}{x-4}.

$$\frac{3\left(x+4\right)\left(-2x+3\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}+\frac{8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Since \frac{3\left(x+4\right)\left(-2x+3\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)} and \frac{8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\left(x+4\right)\left(-2x+3\right)+8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Do the multiplications in 3\left(x+4\right)\left(-2x+3\right)+8\left(x-4\right).

$$\frac{-6x^{2}+9x-24x+36+8x-32}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Combine like terms in -6x^{2}+9x-24x+36+8x-32.

$$\frac{-6x^{2}-7x+4}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Expand \left(x-4\right)\left(x+4\right)\left(-2x+3\right).

$$\frac{-6x^{2}-7x+4}{-2x^{3}+3x^{2}+32x-48}$$

Expand

$$\frac{4-7x-6x^{2}}{\left(3-2x\right)\left(x^{2}-16\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-4 and \left(x+4\right)\left(3-2x\right) is \left(x-4\right)\left(x+4\right)\left(-2x+3\right). Multiply \frac{3}{x-4} times \frac{\left(x+4\right)\left(-2x+3\right)}{\left(x+4\right)\left(-2x+3\right)}. Multiply \frac{8}{\left(x+4\right)\left(3-2x\right)} times \frac{x-4}{x-4}.

$$\frac{3\left(x+4\right)\left(-2x+3\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}+\frac{8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Since \frac{3\left(x+4\right)\left(-2x+3\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)} and \frac{8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\left(x+4\right)\left(-2x+3\right)+8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Do the multiplications in 3\left(x+4\right)\left(-2x+3\right)+8\left(x-4\right).

$$\frac{-6x^{2}+9x-24x+36+8x-32}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Combine like terms in -6x^{2}+9x-24x+36+8x-32.

$$\frac{-6x^{2}-7x+4}{\left(x-4\right)\left(x+4\right)\left(-2x+3\right)}$$

Expand \left(x-4\right)\left(x+4\right)\left(-2x+3\right).

$$\frac{-6x^{2}-7x+4}{-2x^{3}+3x^{2}+32x-48}$$