Algebra expressions

Question

Find three consecutive even integers such that half the sum of the first and third integers is 4 more than twice the second integer.

Answer 1

(F+T)/2 = 4+2S

(x+ x+4)/2 = 4+2(x+2)

2x+ 4 = 2(4+2x+4) = 16+4x

2x = -12

x=-6

x+2 = -4

x+4 = -2

the 3 consecutive even numbers are -6,-4, -2

first & last sum to - 8. half of the sum = -4

twice the 2nd = -8. Add 4 to get -4