There are three consecutive even integers such that 10 less than the sum of the first two is equal to the third. What are the integers?

Answer 1

Hi Justin, the three integers are 8, 10 and 12. The explanation is given below. You can contact me if you need more clarification.

First keep in mind that we are looking for 3 consecutive even integers that satisfy the rules:

10 less than the sum of first two of the even integers = two-thirds the third integer.

Let us represent the first integer with the letter x.

Note that when counting numbers(integers in our case), starting from an even integer say 4, it takes two jumps to get to the next even integer: 4 + 2 = 6(next even integer).

So if the first integer is x, then the second integer should be x + 2.

Similarly, the third integer is (x + 2) + 2 = x+2+2 = x+4. {Since the third even integer, comes two-integer jumps after the second even integer}.

Now let us tackle the rules:

Sum of the first two of the consecutive even integers

= first even integer + second even integer = x + (x+2) = x + x + 2 = 2x + 2

10 less than the sum of the first two of these even integers

= 2x + 2 - 10 = 2x - 8

two-thirds the third integer

= 2/3 * third integer = (2/3) * (x + 4)

Now let us substitute the mathematical formulas for the main rule:

10 less than the sum of first two of the even integers = two-thirds the third integer

2x - 8 = 2/3*(x + 4) (we can now solve this equation for x): first multiply both sides by 3

3*[2x-8] = 3*[2/3*(x + 4)]

6x - 24 = 2*(x + 4)

6x - 24 = 2x + 8

6x - 2x = 8 + 24

4x = 32 (divide both sides by 4)

4x/4 = 32/4

so x = 8

This means that our first even integer is: x = 8

Hence second even integer is : x+2 = x+2 = 8+2 = 10

Lastly, third even integer is: x+4 = 8+4 = 12

So we have our answer: The three integers are 8, 10 and 12

I hope this helps. Good luck!