After you set the expressions equal, move all terms to one side so that the equation equals $0$. What quadratic equation in $x$ do you get?

You do not need each solution separately. For a quadratic equation $ax^2 + bx + c = 0$, recall the simple expression that gives you the sum of its solutions in terms of $a$ and $b$.

Identify $a$ and $b$ in your quadratic and plug them into the formula for the sum of the roots, $x_1 + x_2 = -b/a$.

In Desmos, enter the two equations exactly:

• y = x^2 + 2x - 5
• y = 3x - 1 This will draw the parabola and the line.

Click on each intersection point of the parabola and the line. Desmos will display the coordinates; note the $x$-coordinates of these intersection points.

In a new expression line, type the sum of the two $x$-coordinates you found (for example, x1 + x2 using the actual numbers). The result shown by Desmos is the sum of all possible $x$-values that satisfy the system.

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

Move all terms to one side so the equation equals $0$:

Now we have a quadratic equation in standard form $ax^2 + bx + c = 0$.

For any quadratic equation $ax^2 + bx + c = 0$ with solutions $x_1$ and $x_2$, the sum of the solutions is

In the equation $x^2 - x - 4 = 0$, identify the coefficients:

• $a = 1$
• $b = -1$
• $c = -4$.

We want $x_1 + x_2$, which is $-b/a$.

Apply the formula $x_1 + x_2 = -\dfrac{b}{a}$ using $a = 1$ and $b = -1$:

So, the sum of all possible values of $x$ that satisfy the system is 1.