Look at the $y$-terms: one equation has $2y$ and the other has $-y$. What could you multiply the second equation by so that its $y$-coefficient becomes $-2y$ and cancels with $2y$ when you add the equations?

After you eliminate $y$, you will get an equation with only $x$. Solve that for $x$, then plug that value into either original equation to solve for $y$.

Once you find a candidate value for $y$, substitute it (with the corresponding $x$) back into both original equations to make sure they are both true.

Type y = (20 - 3x)/2 into Desmos. This is the same as $3x + 2y = 20$, solved for $y$.

On a new line, type y = 5x - 3, which is the second equation solved for $y$.

Look for the point where the two lines intersect and tap or click on it; note the $y$-coordinate of that intersection. That $y$-value is the answer to the question.

We are given a system of two equations:

We want the value of $y$ in the solution $(x,y)$. A fast way is elimination: make the coefficients of $y$ opposites so they cancel when we add the equations.

Multiply the second equation by $2$ so that its $y$-term becomes $-2y$:

Second equation: $5x - y = 3$

Multiply by $2$:

Now line this up with the first equation and add them:

When you add these, the $2y$ and $-2y$ cancel.

Add the two equations:

So we have the $x$-value of the solution. Next we use this to find $y$.

Use either original equation. The second one is simpler:

Substitute $x = 2$:

So the solution has $y=7$, which corresponds to answer choice C.