The following rational equation has denominators that contain variables. for this equation, a. write the value or values of the variable that make a denominator zero. these are the restrictions on the variable. b. keeping the restrictions in mind, solve the equation.

also what is/are the value or values of the variable that make(s) the denominators zero?

  a)  x=3 and x=-6 will make the denominator terms zero, hence x may not have those values.

  b)  no solution

The left side can be written over a common denominator so as to make the equation be ...

  ((x+6) -3(x-3))/((x-3)(x+6)) = 9/((x -3)(x +6))

  (-2x +15)/((x -3)(x +6)) = 9/((x -3)(x +6)) . . . . simplify

  -2(x -3)/((x -3)(x +6)) = 0 . . . . . . . . . . . . . . . . subtract the right side

  -2/(x+6) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . cancel factors of (x -3)

This equation has no solution. (There is no value of x that will make -2 = 0.)

Comment on denominator zeros

After you've seen a few instances of the equation

  x + c = 0

it doesn't take long for you to recognize that the solution is ...

  x = -c

That is, when a denominator factor is (x+c), the corresponding value of x that makes this factor zero is x = -c. So, a denominator factor of x+c means the domain must exclude x = -c.

Knowing this, we can write the restrictions on a rational function "by inspection" without having to do anything except mental pattern matching.

Comment on the solution method

By rewriting the rational equation to be something of the form (a/b) = 0, we have the opportunity, as here, to cancel factors that would otherwise result in extraneous solutions. Then, we don't have to be concerned with whether or not the solutions found are extraneous.