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?
$$\dfrac{3}{x}\ +\ \dfrac{4}{y\ }\ =\ 3$$
$$-\dfrac{2}{x}\ +\ \dfrac{1}{y\ }\ =\ -13$$
Solution
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.
