Find the sum:

$(v i)$ $3 x + 4 x y - y^{2}, x y - 4 x + 2 y^{2}$ and $3 y^{2} - x y + 6 x$

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Solution

$Step:$ Find the sum

By grouping the terms having the same literals and adding, we get,

$(3 x + 4 x y - y^{2}) + (x y - 4 x + 2 y^{2}) + (3 y^{2} - x y + 6 x)$

$= 3 x + 4 x y - y^{2} + x y - 4 x + 2 y^{2} + 3 y^{2} - x y + 6 x$

$= (3 x - 4 x + 6 x) + (4 x y + x y - x y) + (- y^{2} + 2 y^{2} + 3 y^{2})$

$= x (3 - 4 + 6) + x y (4 + 1 - 1) + y^{2} (- 1 + 2 + 3)$

$= 5 x + 4 x y + 4 y^{2}$

Hence, the sum of given terms is $5 x + 4 x y + 4 y^{2} .$

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