Question

which expression shows \\(\frac{3x + 7}{x - 5} + \frac{6x}{x + 2}\\) as one fraction?\\(\bigcirc\\) \\(\frac{9x^2 - 17x + 14}{x^2 - 3x - 10}\\)\\(\bigcirc\\) \\(\frac{9x^2 - 16x + 13}{x^2 - 3x - 10}\\)\\(\bigcirc\\) \\(\frac{9x^2 + 31x + 14}{x^2 - 3x - 10}\\)\\(\bigcirc\\) \\(\frac{9x^2 + 43x + 14}{x^2 - 3x - 10}\\)

Explanation:

Step1: Find common denominator

The common denominator of $\frac{3x+7}{x-5}$ and $\frac{6x}{x+2}$ is $(x-5)(x+2)$. Expand it:
$$(x-5)(x+2)=x^2+2x-5x-10=x^2-3x-10$$

Step2: Rewrite fractions with common denominator

Adjust each fraction to have the common denominator:
$$\frac{(3x+7)(x+2)}{(x-5)(x+2)} + \frac{6x(x-5)}{(x-5)(x+2)}$$

Step3: Expand numerators

Expand $(3x+7)(x+2)$ and $6x(x-5)$:
$$(3x+7)(x+2)=3x^2+6x+7x+14=3x^2+13x+14$$
$$6x(x-5)=6x^2-30x$$

Step4: Add numerators together

Combine the expanded numerators over the common denominator:
$$\frac{3x^2+13x+14 + 6x^2-30x}{x^2-3x-10}$$
$$=\frac{(3x^2+6x^2)+(13x-30x)+14}{x^2-3x-10}$$
$$=\frac{9x^2-17x+14}{x^2-3x-10}$$

Answer:

$\boldsymbol{\frac{9x^2 - 17x + 14}{x^2 - 3x - 10}}$ (the first option)