# How do you add (2x^2 - 4xy + y^2) + ( -6x^2 - 9xy - y^2) + (x^2 + xy - 6y^2)?

May 11, 2018

$- 3 {x}^{2} - 12 x y - 6 {y}^{2}$

#### Explanation:

When adding terms in polynomials, you are allowed to combine "like terms".

Two terms are "like" if they have the same variable(s) with possibly different constant coefficients. Recall that in a term like $17 {x}^{2}$, the number $17$ is the coefficient and the ${x}^{2}$ is the variable.

For some examples, $6 {x}^{2}$ and $8 {x}^{2}$ are "like terms", because their variable parts are both ${x}^{2}$. Also, $x y z {h}^{2}$ and $- 13 x y z {h}^{2}$ are also like, since their variables are both $x y z {h}^{2}$. The terms $3 y$ and $3 x y$ are not like terms, because one has a variable of $x$ and the other of $x y$.

If you have two "like terms", you add them (or subtract them) by adding (or subtracting) their coefficients. So $2 {x}^{2} + 5 {x}^{2} = 7 {x}^{2}$. Similarly, $4 x y - 7 x y = - 3 x y$.

Now to the problem at hand. We have:

$2 {x}^{2} - 4 x y + {y}^{2} - 6 {x}^{2} - 9 x y - {y}^{2} + {x}^{2} + x y - 6 {y}^{2}$

To make things more clear, we'll rearrange the terms (keeping their positive or negative signs) so that like terms are next to one another.

$\left(2 {x}^{2} - 6 {x}^{2} + {x}^{2}\right) + \left(- 4 x y - 9 x y + x y\right) + \left({y}^{2} - {y}^{2} - 6 {y}^{2}\right)$

Now we just add the coefficients.

$\left(2 - 6 + 1\right) {x}^{2} + \left(- 4 - 9 + 1\right) x y + \left(1 - 1 - 6\right) {y}^{2}$
$= - 3 {x}^{2} - 12 x y - 6 {y}^{2}$

Now there are no two like terms, meaning we are done adding. This is our final answer.