Solve (-5(x-3)/2x^2+3x-20)+((16x-(2(x-8)-5))/2x^2+3x-20)

$$( \frac{ -5(x-3) }{ 2 { x }^{ 2 } +3x-20 } )+( \frac{ (16x-(2(x-8)-5)) }{ 2 { x }^{ 2 } +3x-20 } )$$

Evaluate

$$\frac{9}{2x-5}$$

Short Solution Steps

To find the opposite of 2\left(x-8\right)-5, find the opposite of each term.

$$\frac{-5\left(x-3\right)}{2x^{2}+3x-20}+\frac{16x-2\left(x-8\right)+5}{2x^{2}+3x-20}$$

Since \frac{-5\left(x-3\right)}{2x^{2}+3x-20} and \frac{16x-2\left(x-8\right)+5}{2x^{2}+3x-20} have the same denominator, add them by adding their numerators.

$$\frac{-5\left(x-3\right)+16x-2\left(x-8\right)+5}{2x^{2}+3x-20}$$

Do the multiplications in -5\left(x-3\right)+16x-2\left(x-8\right)+5.

$$\frac{-5x+15+16x-2x+16+5}{2x^{2}+3x-20}$$

Combine like terms in -5x+15+16x-2x+16+5.

$$\frac{9x+36}{2x^{2}+3x-20}$$

Factor the expressions that are not already factored in \frac{9x+36}{2x^{2}+3x-20}.

$$\frac{9\left(x+4\right)}{\left(2x-5\right)\left(x+4\right)}$$

Cancel out x+4 in both numerator and denominator.

$$\frac{9}{2x-5}$$

Expand

$$\frac{9}{2x-5}$$

Short Solution Steps

To find the opposite of 2\left(x-8\right)-5, find the opposite of each term.

$$\frac{-5\left(x-3\right)}{2x^{2}+3x-20}+\frac{16x-2\left(x-8\right)+5}{2x^{2}+3x-20}$$

Since \frac{-5\left(x-3\right)}{2x^{2}+3x-20} and \frac{16x-2\left(x-8\right)+5}{2x^{2}+3x-20} have the same denominator, add them by adding their numerators.

$$\frac{-5\left(x-3\right)+16x-2\left(x-8\right)+5}{2x^{2}+3x-20}$$

Do the multiplications in -5\left(x-3\right)+16x-2\left(x-8\right)+5.

$$\frac{-5x+15+16x-2x+16+5}{2x^{2}+3x-20}$$

Combine like terms in -5x+15+16x-2x+16+5.

$$\frac{9x+36}{2x^{2}+3x-20}$$

Factor the expressions that are not already factored in \frac{9x+36}{2x^{2}+3x-20}.

$$\frac{9\left(x+4\right)}{\left(2x-5\right)\left(x+4\right)}$$

Cancel out x+4 in both numerator and denominator.

$$\frac{9}{2x-5}$$