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# The sum of five consecutive even integers is ten more than four times the middle number. Find the integers.

I'm really sorry I made a mistake but, again how do you solve this question and can you please explain?

### 2 Answers by Expert Tutors

Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
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let's say x is the first of the numbers, so, the next ones are (x +2), (x + 4), (x + 6), ( x +8).

here they are again: x, (x +2), (x + 4), (x + 6), ( x +8). We can tell here that (x + 4) is the middle number,

so, here's the equation:

x + (x +2) + (x +4) + (x + 6) + (x + 8) = 10 + 4(x + 4)

5x + 20 = 10 + 4x + 16

5x + 20 = 4x + 26

subtract 20 from both sides:

5x + 20 -20 = 4x + 26 -20 ----> 5x = 4x + 6

now, subtract 4x from both sides:

5x - 4x = 4x - 4x + 6

so x = 6.

so, the consecutive even integers are:  6, 8, 10, 12, 14

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
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Hi again Salma;
No need to apologize.  I enjoy this.
x
x+2
x+4--middle
x+6
x+8
TOTAL...5x+20
5x+20=[4(x+4)]+10
Let's distribute...
5x+20=4x+16+10
Let's combine like terms...
5x+20=4x+26
Let's subtract 4x from both sides...
-4x+5x+20=4x+26-4x
x+20=26
Let's subtract 20 from both sides...
-20+x+20=26-20
x=6

x=6
x+2=8
x+4=10---middle
x+6=12
x+8=14
TOTAL...50

50=[4(10)]+10