Solve each inequality for $x$ in terms of $y$. This will show the range of $x$-values allowed for a given $y$.

Once you have an upper and lower bound for $x$ in terms of $y$, ask what condition those bounds must satisfy for some $x$ to lie between them. Then solve the resulting inequality in $y$.

In Desmos, enter $2x+3y=9$ and $x-y=1$ as separate equations. These graphs are the boundary lines for the inequalities.

Click the point where the two lines intersect. Its $y$-coordinate is the upper boundary for the possible $y$-values, although that boundary is not included because the first inequality is strict.

Enter $2x+3y<9$ and $x-y\ge1$. Desmos shades the solution region. Verify that its $y$-values extend below, but not to, the intersection point's $y$-coordinate; then select the matching answer choice.

We want all possible $y$-coordinates for points $(x,y)$ that satisfy both inequalities.

Start by solving each inequality for $x$ in terms of $y$:

From $2x+3y<9$:

From $x-y\ge 1$:

So, for a given $y$, any valid $x$ must satisfy both $x\ge 1+y$ and $x<\frac{9-3y}{2}$.

For a $y$-value to be allowed, there must be at least one $x$ that fits between the lower and upper bounds.

The lower bound on $x$ is $1+y$ from $x-y\ge 1$.

The upper bound on $x$ is $\frac{9-3y}{2}$ from $2x+3y<9$.

For there to be any $x$ at all, the lower bound must be strictly less than the upper bound:

The inequality must be strict because $x$ has to be at least the lower bound and less than the upper bound. If the bounds were equal, no value of $x$ could satisfy both conditions.

Now solve

First clear the fraction by multiplying both sides by $2$:

Therefore, the possible $y$-coordinates are exactly the real numbers satisfying $y<\frac{7}{5}$, which is choice D.