John has two lists of consecutive numbers. The first two numbers of each list equal ten. If the third number in the second list times 2 plus 8 = the sum of the first list, what are the numbers in the two lists? The answer is 6--8-10 and 4-6-8. Need help understanding how to set up the equation.

The first list of even number is ** 2n** , 2n + 2 , (2n + 2) + 2

The second list is (

*) , (10 - 2n) + 2 , ((10 - 2n) + 2) + 2*

**10 - 2n**↑ ↑

*because "the sum of first two numbers of each list equal ten"*↑

↑

the third number in the second list

2{((10 - 2n) + 2) + 2} + 8 = 2n + (2n + 2) + ((2n + 2) +2)

2 (10 - 2n + 4) + 8 = 6n + 6

20 - 4n + 8 + 8 = 6n + 6

36 - 4n = 6n + 6

-4n - 6n = -36 + 6

-10n = -30 ----> n = 3

First list: 2 × 3 =

**6**

**8**

**10**

Second list: 10 - 6 =

**4**

**6**

**8**

## Comments

Judy, according to the answer you gave us, you missed very important word: "John has two lists of consecutive

evennumbers."