$\frac{x^{2}+18 x+81}{x^{2}-49} \cdot \frac{x+7}{x+9}$

A) $\frac{x+9}{x+7}$

B) $\frac{x+9}{x-7}$

C) $\frac{x+9}{(x-7)^{2}}$

D) $\frac{x-9}{x-7}$

Question

$\frac{x^{2}+18 x+81}{x^{2}-49} \cdot \frac{x+7}{x+9}$

A) $\frac{x+9}{x+7}$

B) $\frac{x+9}{x-7}$

C) $\frac{x+9}{(x-7)^{2}}$

D) $\frac{x-9}{x-7}$

Quizplus · Accepted Answer

The expression simplifies by factoring the numerator and the denominator of the first fraction as $(x+9)^2$ and $(x+7)(x-7)$, respectively. After canceling out $(x+9)$ from the numerator of the first fraction and the denominator of the second fraction, the simplified form is $\frac{x+9}{x-7}$.

x2+18x+81x2−49⋅x+7x+9\frac{x^{2}+18 x+81}{x^{2}-49} \cdot \frac{x+7}{x+9}x2−49x2+18x+81​⋅x+9x+7​ A) x+9x+7\frac{x+9}{x+7}x+7x+9​ B) x+9x−7\frac{x+9}{x-7}x−7x+9​ C) x+9(x−7)2\frac{x+9}{(x-7)^{2}}(x−7)2x+9​ D) x−9x−7\frac{x-9}{x-7}x−7x−9​

$\frac{x^{2}+18 x+81}{x^{2}-49} \cdot \frac{x+7}{x+9}$ A) $\frac{x+9}{x+7}$ B) $\frac{x+9}{x-7}$ C) $\frac{x+9}{(x-7)^{2}}$ D) $\frac{x-9}{x-7}$