Answer
6
Work Step by Step
For cross multiplication, we know:
$$\frac{a}{d} =\frac{b}{c} \rightarrow ac=bd$$
Thus, we find:
$$9\left(x-3\right)=\left(x^2-6x+9\right)\cdot \:3 \\ x=3,\:x=6$$
Checking our solution by plugging the given value back into the equation, we assure that the solution $x=3$ IS extraneous.
