solve this for me three times the greatest of three consecutive?

Question

even integers exceeds twice the least by 38. what are the integers?

Answer 1

I won't solve it for you, but I'll show you how to do it.

Let the smallest number be 2n (that makes it even).
Then the others are 2n+2 and 2n+4 (because they are consecutive).

The question therefore states that
3(2n+4) = 2*2n + 38

Simply solve for n, from which you can get the numbers (2n, 2n+2, 2n+4).

Answer 2

So, you have three consecutive, even integers, like 2, 4, and 6. But instead, we'll call them x, x+2, and x+4.

So, we can make an equation for this problem. The greatest integer, x+4, is multiplied by 3. We can write that like this:

3(x+4) =

And that exceeds twice the least (2x) by 38. We can write the rest like this:

3(x+4) = 2x + 38

Now, solve the equation for x, knowing that x is the least integer.

3x + 12 = 2x + 38

x = 26
x+2 = 28
x+4 = 30

And there's your answer! Go ahead and check it:

three times the greatest (30x 3 = 90) exceeds twice the least (2x = 52) by 38.

90-52=38

So, the answer is 26, 28, and 30.

Answer 3

3(x+4) = 2x + 38
3x + 12 = 2x + 38
x = 26

So the integers are 26, 28 and 30

Answer 4

3(2x + 4) = 2(2x) + 38
=> 2x = 26
=> x = 13
numbers are 2x, 2x + 2, 2x + 4
or 2(13), 2(13) + 2, 2(13) + 4
26, 28 and 30

Answer 5

your variables: x, x+2, x+4
your equation: 3(x+4) = 2x + 38