Question

solve the equation $\frac{1}{x-3}=\frac{2}{x^2-6x+9}$.

Explanation:

Step1: Factor the right denominator

Notice that $x^2-6x+9=(x-3)^2$, so the equation becomes:
$$\frac{1}{x-3}=\frac{2}{(x-3)^2}$$

Step2: Eliminate denominators

Multiply both sides by $(x-3)^2$ (note $x
eq3$ to avoid division by zero):
$$(x-3)^2 \cdot \frac{1}{x-3}=(x-3)^2 \cdot \frac{2}{(x-3)^2}$$
Simplify to get:
$$x-3=2$$

Step3: Solve for x

Add 3 to both sides:
$$x=2+3$$

Answer:

$x=5$