Answer
$x=4; y=3$ or $(4, 3)$
Work Step by Step
We need to solve the given system of equations:
$$\dfrac{4}{x}-\dfrac{3}{y}=0 \quad \quad \quad(1)\\
\dfrac{6}{x}+\dfrac{3}{2y}=2\quad \quad \quad(2)$$
First, we will multiply equation $(1)$ by $\frac{1}{2}$ to get
$$\dfrac{4}{2x}-\dfrac{3}{2y}=0\quad \quad \quad(3)$$
Now, add equations $(2)$ and $(3)$ to obtain:
\begin{align*}
\\\left(\dfrac{6}{x}+\dfrac{3}{2y}\right)+\left(\dfrac{4}{2x}-\dfrac{3}{2y}\right)&=2+0\\
\\\dfrac{6}{x}+\dfrac{4}{2x}&=2\\
\\\dfrac{6}{x}+\dfrac{2}{x}&=2\\
\\\dfrac{8}{x}&=2\\
\\8&=2(x)\\
\\4&=x
\end{align*}
Finally, back substitute $x$ into Equation $(1)$ to solve for $y$:
$$\dfrac{4}{4} -\dfrac{3}{y}=0 \implies 1-\dfrac{3}{y}=0 \implies 1=\dfrac{3}{y} \implies y=3$$
So the solution is $x=4; y=3$ or $(4, 3)$.
