Question

find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to 742​

Answer 1

Here let the four even integers be x, x+2, x+4, x+6 then the equation as per the question is..

x + x + 2 + 2 ( x + 4 + x + 6 ) = 742

2x + 2 + 2 ( 2x + 10 ) = 742 [2×2x = 4x ; 2×10 = 20]

2x + 2 + 4x + 20 = 742

6x + 22 = 742

6x = 742 - 22

6x = 720

x = 720 ÷ 6 [ 720 ÷ 6 = 120 ]

Therefore, x = 120

Then the four consecutive integers are,

x = 120

x + 2 = [ 120 + 2 ] = 122

x + 4 = [ 120 + 4 ] = 124

x + 6 = [ 120 + 6 ] = 126

Answer 2

Solution :

Let, the first consecutive even integer be, x.

Then, the second consecutive even integer be, x + 2.

So, the third consecutive even integer be, x + 4.

And, the fourth consecutive even integer be, x + 6.

It is Given that,

The sum of the first two added to twice the sum of the last two is equal to 742.

As per question, we get,

➮ x + x + 2 + 2(x + 4 + x + 6) = 742

➮ x + x + 2 + 2x + 8 + 2x + 12 = 742

➮ x + x + 2x + 2x + 2 + 8 + 12 = 742

➮ 6x + 22 = 742

➮ 6x = 742 - 22

➮ 6x = 720

➮ x = 720/6

➮ x = 120.

Therefore, We got the value of, x = 120.

Hence,

The first consecutive even integer, x = 120.

Then, the second consecutive even integer , x + 2 = 120 + 2 = 122.

So, the third consecutive even integer, x + 4 = 120 + 4 = 124.

And, the fourth consecutive even integer, x + 6 = 120 + 6 = 126.