Solve 4/x-3-5x/x+4=

$$\frac { 4 } { x - 3 } - \frac { 5 x } { x + 4 } =$$

Evaluate

$$\frac{16+19x-5x^{2}}{\left(x-3\right)\left(x+4\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and x+4 is \left(x-3\right)\left(x+4\right). Multiply \frac{4}{x-3} times \frac{x+4}{x+4}. Multiply \frac{5x}{x+4} times \frac{x-3}{x-3}.

$$\frac{4\left(x+4\right)}{\left(x-3\right)\left(x+4\right)}-\frac{5x\left(x-3\right)}{\left(x-3\right)\left(x+4\right)}$$

Since \frac{4\left(x+4\right)}{\left(x-3\right)\left(x+4\right)} and \frac{5x\left(x-3\right)}{\left(x-3\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\left(x+4\right)-5x\left(x-3\right)}{\left(x-3\right)\left(x+4\right)}$$

Do the multiplications in 4\left(x+4\right)-5x\left(x-3\right).

$$\frac{4x+16-5x^{2}+15x}{\left(x-3\right)\left(x+4\right)}$$

Combine like terms in 4x+16-5x^{2}+15x.

$$\frac{19x+16-5x^{2}}{\left(x-3\right)\left(x+4\right)}$$

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

$$\frac{19x+16-5x^{2}}{x^{2}+x-12}$$

Differentiate w.r.t. x

$$\frac{4\left(-6x^{2}+22x-61\right)}{\left(\left(x-3\right)\left(x+4\right)\right)^{2}}$$