If you solve the linear equation $3x + 2y = 10$ for $x$ and substitute into $y = 2x^2 - 5x + 4$, you will get an equation that involves only $y$. Put it into standard quadratic form.

Once you have a quadratic equation in $y$ of the form $ay^2 + by + c = 0$, think about how the roots $y_1$ and $y_2$ relate to $a$, $b$, and $c$ without actually solving for each root.

Remember that for a quadratic $ay^2 + by + c = 0$, there are simple formulas for the sum and product of the roots. Which one gives you $y_1y_2$ directly?

Enter the two equations into Desmos:

• Type y = 2x^2 - 5x + 4.
• Rewrite the line as $y = \dfrac{10 - 3x}{2}$ and type y = (10 - 3x)/2. Desmos will show the intersection points of the parabola and the line.

Click on each intersection point where the graphs meet. Desmos will display the coordinates $(x_1, y_1)$ and $(x_2, y_2)$. Note the two $y$-values from these points.

In a new Desmos expression line, type the product of the two $y$-values you observed (for example, y1 * y2 using the numerical values Desmos showed). The resulting value is $y_1y_2$; compare this exact value with the answer choices.

From the linear equation $3x + 2y = 10$, solve for $x$ in terms of $y$:

$3x = 10 - 2y \Rightarrow x = \dfrac{10 - 2y}{3}$.

Substitute this expression for $x$ into the quadratic equation $y = 2x^2 - 5x + 4$:

Now the equation has only the variable $y$.

First square and distribute carefully:

so

Also,

Write everything over a common denominator of $9$:

Combine the numerators:

So

Multiply both sides by $9$ and move all terms to one side:

Thus $y_1$ and $y_2$ are the two roots of the quadratic

For any quadratic equation of the form

the product of its two roots is

Here, $a = 8$ and $c = 86$ in the equation $8y^2 - 59y + 86 = 0$, so

Now we only need to simplify this fraction and match it to a choice.

Simplify $\dfrac{86}{8}$ by dividing numerator and denominator by $2$:

Therefore,

which corresponds to choice C.