4. Forty-eight more than the second of three consecutive odd integers is the sum of the first Ang 2/3 of the third the three integers.

Answer 1

Let the three consecutive odd integers be x, x+2, and x+4.

Given that forty-eight more than the second integer is equal to the sum of the first and 2/3 of the third integer:

x + 2 + 48 = x + (2/3)(x+4)

Subtract x from both sides:

2 + 48 = (1 + 2/3)x + (8/3)

50 = (3/3 + 2/3)x + (8/3)

50 = (5/3)x + (8/3)

Multiply both sides by 3 to clear the fraction:

150 = 5x + 8

Subtract 8 from both sides:

142 = 5x

Divide by 5:

x = 28

So the three consecutive odd integers are 28, 30, and 32.

Therefore, forty-eight more than the second integer (30) is 78, and 2/3 of the third integer (32) is 64/3. These two values are indeed equal.