The first equation has $-3y$ and the second has $+y$. What could you multiply the second equation by so that its $y$-term becomes $+3y$ and will cancel with $-3y$ when you add the equations?

After you eliminate $y$, solve the resulting equation for $x$. Then plug that value of $x$ into one of the original equations (pick the simpler one) to solve for $y$.

When you substitute $x$ into an equation, watch the negative sign on the right-hand side and combine the fractions carefully so you do not lose a minus sign.

Rewrite each equation in terms of $y$:

• From $2x - 3y = 7$, solve for $y$: $y = \frac{2x - 7}{3}$.
• From $4x + y = -1$, solve for $y$: $y = -1 - 4x$.

In Desmos, type these as two separate lines:

• y = (2x - 7)/3
• y = -1 - 4x

Look at the graph and tap (or click) the point where the two lines intersect. Desmos will display the intersection as an ordered pair $(x, y)$; the $y$-coordinate of this point is the solution for $y$ in the system.

We have the system

To eliminate $y$, make the coefficients of $y$ opposites. The first equation has $-3y$, so multiply the entire second equation by $3$ to get a $+3y$ term:

Now add the new equation to the first equation so that $y$ cancels:

Adding the equations gives:

Solve for $x$:

Use one of the original equations to find $y$. The second equation is simpler:

Substitute $x = \frac{2}{7}$:

Simplify $4\left(\frac{2}{7}\right)$:

From

subtract $\frac{8}{7}$ from both sides:

Write $-1$ as $-\frac{7}{7}$ so the denominators match:

So the value of $y$ is $-\frac{15}{7}$, which corresponds to choice D.