Question

Find 3 consecutive interferes such that the sum of the first and twice the second is 62 minus three times the third

Answer 1

let x be the first number,

so, the next two are (x+ 1) and (x+2)

x + 2(x +1) = 62 - 3(x + 2)

x + 2x + 2 = 62 -3x - 6

3x + 2 = 62 - 3x - 6

3x + 3x = 62 - 6 - 2

6x = 54

x = 9

so, three consecutive integers are

9, 10, 11

Hope this helps!