Once you have $y$ written in terms of $x$, plug that expression into the equation $xy = 12$ to eliminate $y$.

After substitution, you will get an equation in $x$ only. Rearrange it into the standard quadratic form $ax^2 + bx + c = 0$ and solve, for example by factoring.

You will get two values for $x$. Make sure you select the one that matches the question: it asks for the smaller possible value of $x$.

From $2x + y = 10$, write $y = 10 - 2x$. From $xy = 12$ (assuming $x \neq 0$), write $y = 12/x$.

In Desmos, enter the two equations as

• y = 10 - 2x
• y = 12/x Make sure both graphs are visible on the same coordinate plane.

Use the Desmos cursor to click on the intersection points of the line and the curve. Note the $x$-coordinates of these intersection points; these are the possible $x$-values that satisfy both equations.

Among the $x$-coordinates of the intersection points you see, identify the smaller one. That is the value you should enter as your answer.

From the first equation,

solve for $y$ in terms of $x$:

This expression can now be substituted into the second equation.

The second equation is

Substitute $y = 10 - 2x$ into this equation:

Now you have an equation with only $x$.

Expand and rearrange:

Move all terms to one side:

Divide by $-2$ to simplify:

Factor the quadratic:

So there are two possible values for $x$.

From $(x - 2)(x - 3) = 0$, the solutions are $x = 2$ and $x = 3$. The question asks for the smaller possible value of $x$, so the correct answer is $2$.