find four consecutive even integers such that the sum of the first three exceeds the fourth by eight.?

Answer 1

Let the numbers be 2x-2, 2x, 2x + 2, 2x + 4
we need the sum of first 3 more than 4th by 8
thus 2x - 2 + 2x + 2x + 2 = 2x + 4 + 8
6x = 2x + 12
4x = 12
x = 3
the numbers are
4, 6, 8, 10

Or you can let the numbers be x - 2, x, x + 2, x + 4
the equation to solve is
3x = x + 12
2x = 12
x = 6.
That is the second number is 6.
The first is 4, third is 8 and 4th is 10.

Answer 2

Let us call the "smallest" integer "n". So the three consecutive integers are: n, n+1, n+2 From what you have told: 3(n+2) = n + n + 1 + 47 3n + 6 = 2n + 48 therefore n = 42 the three integers are therefore: 42, 43, 44

Answer 3

Let the integers be: (a - 3d), (a - d), (a + d) and (a + 3d), then the sum of the first three = (a - 3d) + (a - d) + (a + d) = 3a - 3d.

It is greater than the fourth by 8, so that: 3a - 3d - (a + 3d) = 8

or 2a - 6d = 8 i.e. a - 3d = 4 i.e. the first integer is 4. Hence next ones are 6, 8 and 10.

So that all the four consecutive even integers are:

4, 6, 8 and 10.

Answer 4

x = "the first integer"
(x + 2) = "the second integer"
(x + 4) = "the third integer"
(x + 6) = "the fourth integer"

x + (x + 2) + (x + 4) = (x + 6) + 8
==> combine like terms
3x + 6 = x + 14
==> subtract x from both sides
2x + 6 = 14
==> subtract 6 from both sides
2x = 8
==> divide by 2 on both sides
x = 4 ... THE FIRST INTEGER IS 4.

So, your four consecutive even integers are 4, 6, 8, and 10.

Answer 5

Let x = 1st; x + 2 = 2nd, x + 4 = 3rd, x + 6 = 4th

Equation:
x + x + 2 + x + 4 = x + 6 + 8
3x + 6 = x + 14
3x - x = 14 - 6
2x = 8
x = 4

Answer: 1st = 4; 2nd = 6; 3rd = 8; 4th = 10

Proof:
4 + 6 + 8 = 10 + 8
18 = 18

Answer 6

let the 4 consecutive even integers be x, x+2, x+4 & x+6
x+x+2+x+4-(x+6)=8
3x+6-x-6=8
2x=8
x=4

first integer = x = 4
second integer = x+2 = 6
third integer = x+4 = 8
fourth integer = x+6 = 10

:):)

Answer 7

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

Four consecutive even integers.... the first three of which equal the fourth.. plus another 8. Then just solve for x.

3x +6 =x +14
2x = 8
x=4

Answer: 4,6,8,10

Answer 8

this becomes AP series with difference of 2
so numbers are a, a+2, a+4, a+6

sum of n terms of AP series with first term a and difference d ( here 2)= n/2 [2a+(n-1)d]

sum of 3 terms = 3/2 [2a+(3-1)2] = 3a+6

now 3a+6 = a+6+8
2a = 8
a= 4
so numbers are 4,6,8,10

Answer 9

the numbers are x ,x+2 ,x+4,x+6
x+x+2+x+4-(x+6)=8
3x+6-x-6=8
2x=8
x=8/2=4
the numbers are 4,6,8,10

Answer 10

2x + (2x + 2) + (2x + 4) = (2x + 6) + 8
6x + 6 = 2x + 14
4x = 8
x = 2
Integers are 4 , 6 , 8 , 10