Answer
$(x,y) =(3,\dfrac{8}{3})$; unique solution
Work Step by Step
Since, we have two equations $\dfrac{x}{8}+\dfrac{3y}{4}=\dfrac{19}{8}$ and $\dfrac{-x}{2}+\dfrac{3y}{4}=\dfrac{1}{2}$
After adding the given two equations, we have $x+4x=19-4$
or, $5x=15$
or, $x=3$
Now, from equation $\dfrac{-3}{2}+\dfrac{3y}{4}=\dfrac{1}{2}$
or,$\dfrac{3y}{4}=\dfrac{1}{2}+\dfrac{3}{2}$
or, $y=\dfrac{8}{3}$
Thus, $(x,y) =(3,\dfrac{8}{3})$; unique solution
