Find three consecutive even integers such that six times the sum of the first and the third is twenty - four greater than eleven times the second .

The three consecutive even numbers are 22, 24 and 26

An even number is a number that can be divided by 2 and there would be no remainder. Examples of even numbers are 2, 4, 6, 8, 10

Let:

x represent the first even number

(x +2) represent the second even number

(x +4) represent the third even number

The equation that represents the verbal explanation in the question is:

6[ x + (x +4)] = 24 + 11(x + 2)

6(2x +4) = 24 + 11x + 22

12x + 24 = 46 + 11x

Combine similar terms

12x - 11x = 46 - 24

x = 22

(x + 2)=  22 + 2 = 24

(x + 4) = 22 + 4 = 26

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1st = 22

2nd = 24

3rd = 26

Step-by-step explanation:

1st = x

2nd = x +2

3rd = x + 4

6(x + x + 4) = 24 + 11(x + 2)

6(2x + 4) = 24 + 11x + 22

12x + 24 = 46 + 11x

x + 24 = 46

x = 22

x + 2 = 24

x + 4 = 26

3

Step-by-step explanation:

D(1 , 1) to D'(3, 3)

so x (1 x 3)= 3

    y (1 x 3) = 3

SF = 3

D(1 , 5) to D'(3, 15)

so x (1 x 3)= 3

    y (5 x 3 ) = 15

SF = 3

Same on coordinate points B to B' and C to C'

so SF = 3