find three consecutive intergers such that the sum of the first and third, plus 3, is 50 more than the second number

Answer: 46, 47, 48

Step-by-step explanation:

1st number=x

2nd number = y which = x+1

3rd number=z which = x+2

X + Z + 3 = 50 + Y

Replace Z with X + 2 and Y with X + 1

X + (X + 2) + 3 = 50 + (X + 1)

Combine like terms

2X + 5 = 51 + X

Then solve for x

2X + 5 - 5 - X = 51 + X - X - 5

So X = 46

46,47,48.

Step-by-step explanation:

46 + 48 + 3 = 97

97 - 50 = 47