Since the question asks about the product of $y$-values, try solving the linear equation for $x$ in terms of $y$ and substitute that expression into the quadratic.

After substitution and simplification, you will get a quadratic equation in $y$. Think about how the coefficients of a quadratic relate to the sum and product of its roots.

In Desmos, enter y = x^2 - 4x - 1 on one line and y = 2x + 5 on another line to graph the parabola and the line.

Click on the points where the line and parabola intersect; Desmos will display the coordinates $(x, y)$ of each intersection. Note both of the $y$-values from these points.

In a new Desmos expression line, type the product of the two $y$-values you found (for example, y1 * y2 using the exact numbers Desmos showed). The resulting output is the product of all possible values of $y$ for the system.

You are given

Because we want the product of all possible values of $y$, it is convenient to get an equation that involves only $y$.

From the linear equation $y = 2x + 5$, solve for $x$:

Substitute $x = \dfrac{y - 5}{2}$ into the quadratic equation $y = x^2 - 4x - 1$:

To clear denominators, multiply every term by 4:

Now expand and simplify the right-hand side:

• $(y - 5)^2 = y^2 - 10y + 25$
• $-8(y - 5) = -8y + 40$

So

Bring all terms to one side:

So the possible $y$-values must satisfy the quadratic equation

For any quadratic equation of the form

the product of its two roots (solutions) is $\dfrac{c}{a}$.

In our equation $y^2 - 22y + 61 = 0$:

• $a = 1$
• $b = -22$
• $c = 61$

Therefore, the product of the two possible values of $y$ is

So the correct answer is 61 (choice C).