Find the missing numerator and denominator.
-$\frac{(\mathrm{x}-2)(\mathrm{x}-6)}{(\mathrm{x}-1)(\mathrm{x}-9)} \cdot \frac{?}{?}=\frac{(\mathrm{x}-6)(\mathrm{x}-3)}{(\mathrm{x}-1)(\mathrm{x}-4)}$

A) $\frac{(x+3)(x-9)}{(x+4)(x-2)}$

B) $\frac{(x+3)(x+9)}{(x+4)(x+2)}$

C) $\frac{(x-3)(x-9)}{(x-4)(x-2)}$

D) $\frac{(x-3)(x+9)}{(x-4)(x+2)}$

Question

Find the missing numerator and denominator.
-$\frac{(\mathrm{x}-2)(\mathrm{x}-6)}{(\mathrm{x}-1)(\mathrm{x}-9)} \cdot \frac{?}{?}=\frac{(\mathrm{x}-6)(\mathrm{x}-3)}{(\mathrm{x}-1)(\mathrm{x}-4)}$

A) $\frac{(x+3)(x-9)}{(x+4)(x-2)}$

B) $\frac{(x+3)(x+9)}{(x+4)(x+2)}$

C) $\frac{(x-3)(x-9)}{(x-4)(x-2)}$

D) $\frac{(x-3)(x+9)}{(x-4)(x+2)}$

Quizplus · Accepted Answer

To solve for the missing numerator and denominator, we need to multiply the left side by a fraction that will cancel out the unwanted terms and introduce the desired terms on the right side. The original equation is:$\frac{(x-2)(x-6)}{(x-1)(x-9)} \cdot \frac{?}{?}=\frac{(x-6)(x-3)}{(x-1)(x-4)}$Looking at the right side, we see that we need to introduce a factor of $(x-3)$ in the numerator and a factor of $(x-4)$ in the denominator. We also need to cancel out the $(x-9)$ in the denominator and the $(x-2)$ in the numerator from the left side. Therefore, the missing fraction must have $(x-3)$ in the numerator to introduce that term and $(x-4)$ in the denominator to introduce that term. It must also have $(x-9)$ in the numerator to cancel out the $(x-9)$ in the denominator of the left side and $(x-2)$ in the denominator to cancel out the $(x-2)$ in the numerator of the left side. Thus, the correct missing fraction is $\frac{(x-3)(x-9)}{(x-4)(x-2)}$, which is option C.