Search 75,527 tutors
1 0

# solving inequalities

the sum of three consecutive even integers is at least 80 and at most 90. what is the largest possible combination of these three integers?

The sum of 3 consecutive even integers is at least 80 and at most

Call the first number X. The second will be X+2 (the next even number). The third is (X+2)+2=X+4. So the sum is X+X+2+X+4= 3X+6

So we now have 80 ≤ 3x+6 ≤ 90

80 -6≤ 3X+6-6≤ 90-6

74≤ 3X≤ 84

74/3≤ 3X/3≤ 84/3

24.666≤ X≤ 28

Since we know X is an even integer, the only options we have are 26 and 28.

We already said the X is the smallest number, and the next two are X+2 and X+4. Plugging in the 2 options for x, we have two possible solutions.

1) 26+28+30= 84

2) 28+30+32=90

Since the question asks for the largest combination, the largest starting even integer in the range would do the job.  The last step of comparing the possible combinations is illustrative, but not necessary.

Keep it up, Gene!!! We need competent "feed back" and I like it. You are right here and were right about f(x) and f(x-1) (hope you remember comment about "x"). You push us to think better!! Here, in America, teacher (at least in public schools) is teaching the same subject level or the same grade years after years ... We forgot all connections before and after and don't see how is working our "work" in future ...

We have three consecutive even numbers; the smallest one, for convenience, will be called x. The second number is two larger than the smallest, so it is called x + 2. The third (and largest) number is two greater than the second, so it is (x + 2) + 2 or x + 4.

Now, we are asked to give the largest three consecutive even numbers such that their sum is at most 90. Therefore, we are asked mathematically that x + (x + 2) + (x + 4) ≤ 90.

x + (x + 2) + (x + 4) ≤ 90;

3x + 6 ≤ 90;

3x ≤ 84;

x ≤ 28.

The smallest number is 28, so the other two numbers are 30 and 32. This makes the final answer to your solution {28, 30, 32}.

Arithmetic:

(80 / 3 ≈ 26.7  and  90 / 3 = 30) ===> 26 , 28 and 30

Algebra:

If     1st even number is "x" ,
then 2nd even number will be "x + 2"
and  3rd even number will be "x + 4" [(x + 2) + 2]

80 ≤ x + x + 2 + x + 4 ≤ 90

80 ≤ 3x + 6 ≤ 90
- 6          - 6    - 6

74  ≤  3x  ≤  84
÷ 3    ÷ 3     ÷ 3

26 ≤ x ≤ 28
Thus, the number are  {26, 28, 30}

First you need to find out what the consecutive integers are for the lowest sum 80.  The equation should be written below to represent three consecutive integers, whose sum = 80.

X + (x+1) + (x+2) = 80

X + x + 1 + x + 2 = 80

3x + 3 = 80

3x = 77

X=25.666

So the next integer is 26. The three consecutive integers are 26+27+28 = 81.

Then you would solve for 3x+3=90, which is the greatest sum.

3x + 3 = 90

3x = 87

X=29

So the integer is 29. The three consecutive integers are 29+30+31=90.

In between these two sets of consecutive integers, there are only a few combinations, you can simply list them and then count how many there are.

1. 26 + 27 + 28 = 81

2. 27 + 28 + 29 = 84

3. 28 + 29 + 30 = 87

4. 29 + 30 + 31 = 90

So there are FOUR possible combinations.