Question

Reduce to lowest terms : $\frac{(3x+4y)}{9x^2-16y^2}$

• A. $\frac{1}{(x-4y)}$
• B. $\frac{1}{(3x-y)}$
• C. $\frac{1}{(3x-4y)}$
  
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{(3x-4y)}$

The lowest term $\frac{3x+4y}{9x^2-16y^2}$ is

$=$ $\frac{3x+4y}{(3x{)}^2-(4y{)}^2}$

$=$ $\frac{3x+4y}{(3x+4y)(3x-4y)}$ ....We know $a^2-b^2=(a+b)(a-b)$

$=$ $\frac{1}{3x-4y}$