After you set $x^2 - 5x + 6 = 4x - 2$, move all terms to one side so that the equation equals 0, then combine like terms to make a standard quadratic equation.

You should get a quadratic of the form $x^2 - 9x + 8 = 0$. Factor it by finding two numbers that multiply to $8$ and add to $-9$, then set each factor equal to 0 to find possible values of $x$.

In Desmos, enter the two equations on separate lines:

• y = x^2 - 5x + 6
• y = 4x - 2

You will see a parabola and a straight line.

Zoom or pan as needed so that you can see where the parabola and the line intersect. Click (or tap) on each intersection point; Desmos will display the coordinates. The $x$-coordinates of these intersection points are the possible values of $x$ that satisfy the system. Either $x$-value shown there is a valid answer to the question.

Because both equations equal $y$, any solution must make the right-hand sides equal:

Move everything to one side to get a quadratic equal to 0:

Now you just need to solve $x^2 - 9x + 8 = 0$.

Factor $x^2 - 9x + 8$.

You need two numbers that multiply to $8$ and add to $-9$. Those numbers are $-1$ and $-8$.

So the factored form is:

If $(x - 1)(x - 8) = 0$, then at least one factor must be zero:

• $x - 1 = 0$, so $x = 1$
• $x - 8 = 0$, so $x = 8$

Both of these $x$-values make the original system true. Since the question asks for one possible value of $x$, you can give $x = 1$ as your answer.