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

Question

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

Answer 1

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

 

 

 

Answer 2

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