The larger of two negative consecutive even integers 2t and 2(t-1) is multiplied by 6 and then added to the smaller of the two original integers. Which of the following represents this operation?
Choose the correct choice from the following answers:
A) 6t-2	B) 8t-2	C) 14t-2	D) 3	E) -2-4t^2
larger of 2t-2, 2t (when they are negative) is 2t. the smallest of 2t and 2t-2 when they are positive is 2t-2.   therefore 6* 2t + 2t-2 = 14t-2.   option 14t-2.
<result>C</result>
The larger of two negative consecutive even integers 2t and 2(t-1) is multiplied by 6 and then added to the smaller of the two original integers. Which of the following represents this operation?