3 consecutive odd intagers such as the 1st and 2nd is equal to 6 less than twice the 3rd

Answer 1

3 consecutive odd integers

1st = x

2nd = x+2

3rd = x+4

1st + 2nd = twice the 3rd minus 6

x + x+2 = 2(x+4) -6

2x + 2 = 2x +8-6

2 = 2

1,3, 5 satisfy the condition: 1+3 = 2(5)-6, 4 = 10-6

7,9, 11 also satisfy the condition 7+9 = 16 = 2(11)-6 = 22-6 = 16

any and all 3 consecutive odd integers satisfy the requirement

an infinite number of solutions

UNLESS possibly you meant

x+ x+2 = 2(x+4-6)

then

2x +2 = 2x -4

2 = -4 which is a contradiction and then there are no solutions.

DNE, null set