Find three even consecutive?

Question

Find three even cosecutive integers such that the sum of the smallest and twice the second is 20 more than the third

Answer 1

Three even consecutive integers:

x, x + 2, x + 4

Such that the sum of the smallest and twice the second:

x + 2(x + 2)

is 20 more than the third:

x + 2(x + 2) = x + 4 + 20
x + 2x + 4 = x + 24
2x = 20
x = 10

So the integers are 10, 12, and 14.

Check: 10 + 2(12) = 14 + 20 check!

Answer 2

let the smallest number = x
the middle number would then be equal to = x+2 since the difference bet. any two POSITIVE integers is 2.
then the largest number would be x+2+2= x+4

the equation:

x + 2(x+2) = x + 4 + 20 (simplify them to get:
x + 2x + 4 = x + 24 (combine like terms to get:
3x - x = 24 - 4 (simplify again to get:
2x = 20 (divide both sides by 2 to have:
x = "10"

substitute the values:
x=10;
x+2= 10+2 = "12"
x+4= 10+4 = "14"

check: x=10

x + 2(x+2) = x + 4 + 20
10 + 2(10+2) = 10 + 4 + 20
10 + 2(12) = 34
10 + 24 = 34
34 = 34
checked! since 34 is 20 more than the third number 14..

Answer 3

Start with x, x + 2, x + 4. Then x + 2(x+ 2) = 20 + (x + 4) => x + 2x + 4 = x + 24 => 3x + 4 = x + 24 => 2x = 20 => x = 10, x + 2 = 12, and x + 4 = 14.

Answer 4

To solve this you need to make an equation.

Let ABC be the consecutive numbers

B = A + 2
C = A + 4
Because the numbers are consecutive

A + 2B = C + 20

A + 2(A + 2) = (A + 4) + 20
A + 2A + 4 = A + 24
3A + 4 = A + 24
Put the terms on their own sides by subtracting both sides by A and 4.
2A = 20
Divide by two
A = 10
B = 10 + 2 = 12
C = 10 + 4 = 14
So the consecutive numbers are 10, 12, and 14.

Answer 5

n+2(n+2)=n+4+20
3n+4=n+24
2n=20
n=10
numbers are 10, 12, 14

check
10+2*12=34
14+20=34

Answer 6

n+2(n+2)=n+4+20
n+2n+4=n+24
2n=20
n=10
10,12,14

Answer 7

first number is x
second number is y
third number is z

x+2y=z
y=x+2
z=y+2

use substitution to figure it out... shouldn't be that hard.