Akash has $6 x y + 4 x y^{2}$ chocolates. He wants to distribute them to 2x children. How many chocolates does each of the child get if y = 2?
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Solution

2x children will get $6 x y + 4 x y^{2}$ chocolates.
So, 1 child will get $\frac{6 x y + 4 x (y^{2})}{2 x}$ chocolates.
Factorize $6 x y + 4 x y^{2}$ by using method of common factors.
$2 \times 3 \times x \times y + 2 \times 2 \times x \times y \times y = 2 x (3 y + 2 y^{2})$
So, $\frac{6 x y + 4 x (y^{2})}{2 x}$ = $\frac{2 x (3 y + 2 y^{2})}{2 x}$ = $(3 y + 2 y^{2}$ )
Given, y = 2. Substituting the value of y in $(3 y + 2 y^{2}$ ) gives,
$3 \times 2 + 2 \times 2 \times 2 = 6 + 8 = 14$

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