William W. answered • 10d

Top Algebra Tutor

1. Let "x" be the first (smallest) integer. Then the next consecutive integer would be "x + 1" and the integer after that would be "x + 2". The sum of these must be 33 so:

x + x + 1 + x + 2 = 33

3x + 3 = 33

3x = 30

x = 30/3 = 10

So the integers (smallest to largest) would 10, 11, 12

2. To ensure we get an odd integer we add 1 to a "for sure" even integer. Such a "for sure" even integer would be 2x (for any integer x, 2x will always be even). If "2x" is always even, then 2x + 1 will always be odd. So let "2x + 1" be the first odd integer then the next consecutive odd integer would be (2x + 1) + 2 or "2x + 3". These two integers must add to 92 so:

2x + 1 + 2x + 3 = 92

4x + 4 = 92

4x = 88

x = 88/4 = 22

Remember that the first odd integer is "2x + 1" so it must be 2(22) + 1 = 45. That makes the next consecutive odd integer 47.

Based on problem 2, an even integer would be 2x where x is any integer. The next two consecutive even integer must be "2x + 2" and "2x + 4" so:

2x + 2x + 2 + 2x + 4 = -198

6x + 6 = -198

6x = -204

x = -204/6 = -34

That makes the first even integer 2(-34) = -68. The next would be -66 and the next would be -64