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
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.
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
My typo:
2 a1 = 56
a1 = 56 / 2 = 28
To find four consecutive even integers that satisfy the given condition, let's assign variables to these unknown numbers.
Let's assume the first even integer is x. Since they are consecutive even integers, the second, third, and fourth even integers can be represented as x + 2, x + 4, and x + 6, respectively.
According to the given condition, the sum of the first and third integers, (x + x + 4), when multiplied by 3, is 44 more than 4 times the fourth integer, (4 * (x + 6)) + 44.
So, we can set up the equation as follows:
3(x + x + 4) = 4(x + 6) + 44
Simplifying the equation:
3(2x + 4) = 4x + 24 + 44
6x + 12 = 4x + 68
6x - 4x = 68 - 12
2x = 56
x = 56 / 2
x = 28
Therefore, the first even integer is 28.
The second even integer is x + 2, which is 28 + 2 = 30.
The third even integer is x + 4, which is 28 + 4 = 32.
The fourth even integer is x + 6, which is 28 + 6 = 34.
So, the four consecutive even integers that satisfy the given condition are 28, 30, 32, and 34.