Assume the questions reads: (1/2)x=(1/3)y+1/10 and 6x-4y=k.
So 6x=k+4y and (1/2)x=(k+4y)/12=y/3+1/10; multiply through by 60: 5k+20y=20y+6, so 5k=6, k=6/5. If k is any other value the equation will not be true.
Let's go back to the original equations. 6x-4y=6/5 or 30x-20y=6, 15x-10y=3 and x/2=y/3+1/10 or 15x=10y+3. These two equations are identical which means there are many solutions and there is a dependence of y on x or vice versa. This relationship is illustrated by a linear graph. So the requirement is that k=6/5.