SOLUTION: find two consecutive even integers such that the lesser added to three times the greater gives a sum of 46

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Question 760719: find two consecutive even integers such that the lesser added to three times the greater gives a sum of 46
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Let the smaller even integer be n.
Then the next consecutive even integer is n+2 (because successive even numbers have a difference of 2)
Given that smaller + 3 times the bigger number gives sum of 46. We can write it algebraically as:
n+%2B+3%2A%28n%2B2%29+=+46 ---- equation (1)
Expanding the brackets, we get
n+%2B+3%2An+%2B+6+=+46
Adding like terms
4%2An+%2B+6+=+46
Subtracting 6 from both sides to get rid of the "6" on the left side
4%2An+%2B+6+-+6+=+46+-+6
4%2An+=+40
Dividing both sides by 4
n+=+40%2F4+=+10
So the smaller number is 10 and the next number is 12. The two consecutive even integers are 10 and 12.
To check the answer, substitute n with 10 in equation (1)
10+%2B+3%2A12+=+10+%2B+36+=+46 ---> which is true.