Question 75·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

The restrictions are values that make the original equation undefined. Since division by $0$ is undefined, $x-3=0$ rules out $x=3$, and $x+2=0$ rules out $x=-2$. Keep both restrictions after the fractions disappear.

Type $y=\frac{2}{x-3}+\frac{3}{x+2}$ and $y=1$. Click both intersections. Desmos shows $x\approx-0.162$ and $x\approx6.162$. Each makes the original sides equal, and neither is a restricted value. Use the graph to count the solutions, not their rounded coordinates to calculate an exact product.

Multiply every term by $(x-3)(x+2)$, the least common denominator. Each fraction’s denominator cancels with its matching factor; the $1$ on the right gets multiplied too:

Distribute each factor, keeping the minus sign in $3(x-3)$:

Combine like terms, which have the same power of $x$:

Subtract $5x-5$ from both sides so the quadratic has $0$ on one side:

For $ax^2+bx+c=0$, the product of its roots is $c/a$. The graph showed that both roots are real and allowed. Here $a=1$ and $c=-1$, so:

The product of all real solutions is $-1$. Grid in -1.