Question

$\frac{(3y+12b)(5a-b)}{(y+4b)(25a^2-b^2)}=$

• A. $\frac{3}{\text{5a-b}}$
• B. $\frac{3}{\text{5a+b}}$
• C. $\frac{3}{\text{3a+b}}$
• D. $\frac{5}{\text{5a+b}}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{3}{\text{5a-b}}$
B $\frac{3}{\text{5a+b}}$

Given, the expression is
$\frac{(3y+12b)(5a-b)}{(y+4b)(25a^2-b^2)}.$

The numerator can be written as,
$(3y+12b)(5a-b)$
$=3(y+4b)(5a-b)$. . .(i)

Applying the identity
$a^2-b^2=(a+b)(a-b)$,
in the denominator, we get,
$(y+4b)(5a+b)(5a-b)$. . .(ii)

Hence, from (i) and (ii),

$\frac{(3y+12b)(5a-b)}{(y+4b)(25a^2-b^2)}=\frac{3(y+4b)(5a-b)}{(y+4b)(5a+b)(5a-b)}$

$=\frac{3}{5a+b}.$