Question

$\frac{3}{4}(a+y)[y+a-\frac{1}{3}(y+a-\frac{1}{4}(a+y\left)\right)]$

• A. $(a+y{)}^2$
• B. $\frac{3a}{16}$
• C. $\frac{9}{16}(a+y{)}^2$
• D. 1

Answer 1

Answer: (C)

$\frac{9}{16}(a+y{)}^2$

$\frac{3}{4}(a+y)[y+a-\frac{1}{3}(y+a-\frac{1}{4}(a+y\left)\right)]$
= $\frac{3}{4}(a+y)[y+a-\frac{1}{3}\left(\frac{3}{4}\right(a+y\left)\right)]$
= $\frac{3}{4}(a+y)[y+a-\frac{1}{4}(a+y\left)\right]$
= $\frac{3}{4}(a+y)\left[\frac{3}{4}\right(a+y\left)\right]$
= $\frac{9}{16}(a+y{)}^2$