three times the second of three consecutive even integers is twelve less than twice the sum of the first and third integers find the largest even integer

Question

hree times the second of three consecutive even integers is twelve less than twice the sum of the first and third integers find the largest even integer

Answer 1

An even integer is 2 times some other integer. So we could write an even integer as 2n.

 

Then three CONSECUTIVE even integers would be written as: 2n, 2n + 2, 2n + 4

 

because you have to add two to get to the next even number.

 

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So 3 times the second one of these is 3(2n + 2) = 6n + 6

 

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The sum of the first and third ones are 2n + 2n + 4 = 4n + 4

 

Then twice the sum of these must be twice this or 2(4n + 4) = 8n + 8

 

Then twelve less than this must be 8n + 8 - 12 = 8n - 4

 

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So 3 times the second is 12 less than twice the sum of the other two means to write

 

6n + 6 = 8n - 4

 

Then 10 = 2n

 

So n has to be 5. 

 

Then our even numbers would be 10, 12, 14

Answer 2

Here's what we know:

3b = 2(a + c) - 12

a = b - 2

c = b + 2 (since a, b, and c are consecutive (in a row) even integers)

 

Since we have values for some of our variables, let's plug them in:

3b = 2(b - 2 + b + 2) - 12

3b = 2(2b) - 12

3b = 4b - 12

Subtract 4b from both sides:

-b = -12

multiply by -1:

b = 12

a = 12 - 2 = 10

c = 12 + 2 = 14

 

check:

3 * 12 = 2(10 + 14) - 12

36 = 2 * 24 - 12

36 = 48 - 12

36 = 36