Question

Find four consecutive odd integers

such that twice the smallest
subtracted from three times the
largest is 63.​

Answer 1

Let:
x = smallest integer
x + 2 = second smallest integer
x + 4 = second largest integer
x + 6 = largest integer

Solving:
(3(x + 6)) - 2x = 63
3x + 18 - 2x = 63
x + 18 = 63
x = 45 (smallest integer)

Answer: The integers are 45, 47, 49, and 51.

Check:
(3*51) - (2*45) = 63
153 - 90 = 63
63 = 63
Are the integers consecutive and odd? Yes!