How do you simplify \[{y^2} + 11y - 6y + {y^2}\] ?

Hint: The given equation is an algebraic equation, that is, it contains a combination of numerical values and alphabets, where alphabets represent some unknown quantities. As the unknown variable quantity “y” is raised to a non-negative integer as the power so it is a polynomial equation. The equation contains the terms $ {y^2},\,11y\,and \;6y $ . The square of “y” is the number obtained when y is multiplied with itself; 11y and 6y represent the multiplication of 11 and y and 6 and y respectively.

Complete step-by-step answer:
Simplification of an equation means to write the number more conveniently and easily so that it can make the calculations easier. So, the given equation can be simplified by grouping the like terms and then applying the given arithmetic operation.
\[{y^2} + 11y - 6y + {y^2}\] can also be written as \[{y^2} + {y^2} + 11y - 6y\]
On performing the arithmetic operations, we get –
 $ 2{y^2} + 5y $
Now, y is present in both the terms, so it can be taken as common –
 $ y(2y + 5) $
Hence, the simplified form of \[{y^2} + 11y - 6y + {y^2}\] is $ y(2y + 5) $ .
So, the correct answer is “ $ y(2y + 5) $ ”.

Note: We cannot find the value of y from the given equation, as the equation is not given equal to some other quantity. We have used the distributive property in the question when we took y as common. The distributive property states that when a number is in multiplication with the sum of the difference of two numbers then the equation can be written as that number multiplied with the first number plus/minus the product of that number with the second number or vice versa, that is, $ a(b \pm c) = ab \pm ac $ or $ ab \pm ac = a(b \pm c) $ .