What is the value of 15x + 15y, where x and y are not equal to 0, such that x and y satisfy the equations given below?

$$\dfrac{3}{x}\ +\ \dfrac{4}{y\ }\ =\ 3$$

$$-\dfrac{2}{x}\ +\ \dfrac{1}{y\ }\ =\ -13$$

The equations given are,

$$\dfrac{3}{x}\ +\ \dfrac{4}{y\ }\ =\ 3$$

$$-\dfrac{2}{x}\ +\ \dfrac{1}{y\ }\ =\ -13$$

Let us assume the value of $$\dfrac{1}{x}$$ to be a and $$\dfrac{1}{y}$$ to be b. The above equations become,

3a + 4b = 3 ----(1)

-2a + b = -13 ----(2)

(1) * 2 + (2) * 3, we get,

(3a + 4b)*2 + (-2a + b)*3 = 3 * 2 + (-13) * 3

6a + 8b - 6a + 3b = 6 - 39

11b = -33

b = -3 and a = 5

We get the values x and y as,

$$\dfrac{1}{x}\ =\ 5$$

$$x\ =\ \dfrac{1}{5}$$

$$\dfrac{1}{y}\ =\ -3$$

$$y\ =\ -\dfrac{1}{3}$$

$$x\ +\ y\ =\ \dfrac{1}{5}-\dfrac{1}{3}\ =\ -\dfrac{2}{15}$$

$$15\left(x\ +\ y\right)\ =\ -\dfrac{2}{15}\ \times\ 15\ \ =\ -2$$

The correct answer is option C.