find four consecutive even integers such that if the sum of the first and third is multiplied by 3 the result is 44 more than 4 times fourth

Answer 1

Four consecutive integers:
x, (x+1), (x+2), (x+3)

As per the question,
3*[x + (x + 2)] = 4(x+3) + 44
3(2x + 2) = 4x + 12 + 44
6x + 6 = 4x + 56

Solve for x.

Answer 2

a1 = first even integers

a2 = second even integers = a1 + 2

a3 = third even integers = a1 + 4

a4 = fourth even integers = a1 + 6

The sum of the first and third is multiplied by 3 the result is 44 more than 4 times fourth mean:

( a1 + a3 ) ∙ 3 = 4 ∙ a4 + 44

( a1 + a1 + 4 ) ∙ 3 = 4 ∙ ( a1 + 6 ) + 44

( 2 a1 + 4 ) ∙ 3 = 4 ∙ a1 + 4 ∙ 6 + 44

2 a1 ∙ 3 + 4 ∙ 3 = 4 a1 + 24 + 44

6 a1 + 12 = 4 a1 + 68

6 a1 - 4 a1 = 68 - 12

2 a1 = 46

a1 = 46 / 2 = 28

The numberas are: 28 , 30 , 32 , 34

Proof:

( a1 + a3 ) ∙ 3 = 4 ∙ a4 + 44

( 28 + 32 ) ∙ 3 = 4 ∙ 34 + 44

60 ∙ 3 = 136 + 44

180 = 180

Answer 3

My typo:

2 a1 = 56

a1 = 56 / 2 = 28