**Quote:**

2*x* # (2*x* @ *y*) = 2*x* # (2*x* + 2*y*) = 3(2*x*)² – (2*x* + 2*y*) = 12*x*² – 2*x* – 2*y*

Let's write the calculations in details:

1.

**2***x* # (2*x* @ *y*) =We apply the definition of the operation "@" to 2

*x* and

*y*.

2

*x* @

*y* = (2

*x*) + 2

*y* = 2

*x* + 2

*y*2.

**2***x* # (2*x* @ *y*) = 2*x* # (2*x* + 2*y*) =We use the definition of the operation "#" for 2

*x* and (2

*x* + 2

*y*).

2

*x* # (2

*x* + 2

*y*) = 3(2

*x*)² – (2

*x* + 2

*y*) = 3 × 2² ×

*x*² – 2

*x* – 2

*y*3 × 2² = 3 × 4 = 12Therefore

**2***x* # (2*x* @ *y*) = 2*x* # (2*x* + 2*y*) = 12*x*² – 2*x* – 2*y*