Question

Find three consecutive even interfere such that their sum is 50 more than the largest interfere

Answer 1

Hm. This is an interesting problem.

Let's see if we can express each of these numbers in terms of x and make an algebra equation to help us solve.

Consecutive even numbers increase by 2.

Let's allow x to equal our first number.

Let's allow x + 2 to equal our 2nd number.

Let's allow x + 4 to equal our 3rd number.

When we add those together, their sum need to equal 50 more than the largest integer (our x + 4)

Let's set it up!

x + (x +2) + (x + 4) = (x + 4) + 50

Simplify!

3x + 6 = x + 54

Let's reorganize the left and right sides.

3x - x = 54 - 6

2x = 48

x = 24 (our 1st even number)

Now, let's check our answer!

24 + 26 + 28 = 78

78 - 50 = 28 Our highest integer.

The sum of consecutive integers 24, 26, 28 is 50 more than the highest integer.