If we convert the inequalities into a standard equation. remember this is a MUST be true question

we do have that

\(y-x<1\)

\(y-x=1\)

\(y=x+1\)

Same for \(3y>x+6\)

\(3y=x+6\)

\(y=x/3+2\)

Now to the observation

1) 𝑦-intercept is 1, i.e. (0,1) and slope is positive. and 𝑦-intercept is 2 i.e. (0,2) and slope is positive so we do know that both are in the I quadrant and must be positive BOTH Therefore D and E are suddenly out

2) y=x+1 and in A a> b should follow that x>y but this is NOT true because we do know that x+1 is Y. For example 3=2+1; x can not be > y; So A is out

3) in B a<b which means that x<y this could be true or not. From the second equality we do know that if x=1 we do have that y=1/3+6=6.3 and in this case a<b BUT if we do have that x is 9 we have that y is also 9 and x=y

The only thing that must be true s that both x and y are positive. The only answer choice correct is C