After you multiply through by the common denominator and simplify, move all terms to one side so the equation is equal to $0$. What type of equation do you get?

The question asks for the product of the real solutions. For a quadratic $ax^2+bx+c=0$, think about how the leading coefficient $a$ and the constant term $c$ are related to the product of its two solutions.

In Desmos, type y = 2/(x-3) + 3/(x+2) on one line and y = 1 on another line. These two graphs represent the left and right sides of the equation.

Zoom or adjust the view until you can see where the two graphs intersect. Click on each intersection point; Desmos will display the coordinates. Note the x-coordinates of the intersection points—those are the real solutions to the equation.

In a new expression line, type the two x-values you found in parentheses, multiplied together (for example, (x_1)*(x_2) using the numerical values you read from the graph). The resulting value shown by Desmos is the product of all real solutions that the question is asking for.

Start with the equation

The common denominator of $x-3$ and $x+2$ is $(x-3)(x+2)$. Multiply every term on both sides by $(x-3)(x+2)$:

This simplifies to

Expand both sides:

• Left side: $2(x+2) = 2x+4$ and $3(x-3) = 3x-9$, so

• Right side: $(x-3)(x+2) = x^2 - x - 6$.

So the equation becomes

Move all terms to one side:

So we have the quadratic

Let the real solutions to $x^2 - 6x - 1 = 0$ be $r_1$ and $r_2$.

For any quadratic of the form

with solutions $r_1$ and $r_2$, a standard result (Vieta's formulas) tells us that

We will use this relationship to find the product of the real solutions without actually solving for $r_1$ and $r_2$ explicitly.

In our quadratic $x^2 - 6x - 1 = 0$:

• $a = 1`
• $c = -1`

So the product of the real solutions is

Therefore, the product of all real solutions to the original equation is $-1$.