Asked by guitarpick03
                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
            
            
        Answers
                    Answered by
            Arora
            
    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.
    
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.
                    Answered by
            Bosnian
            
    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
    
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
                    Answered by
            Bosnian
            
    My typo:
2 a1 = 56
a1 = 56 / 2 = 28
    
2 a1 = 56
a1 = 56 / 2 = 28
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