Question

Question !!

find three consecutive even integer such that the sum of twice the first integer four more than the second and seven less that the third integer is 27 find the integers ?

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Answer 1

Answer :

• Required integers are 6, 8 & 10.

Given that :-

• Sum of twice the first integer four more than the second integer and seven less than the third integer is 27.

To Find :-

• What are the three integers ?

Solution :-

• Let the first integer be x
• So, second integer is (x + 2)
• Also, Third integer is (x + 2 + 2) = (x + 4)

★ According to the question :

• Sum of twice the first integer four more than the second integer and seven less than the third integer is 27.

Therefore,

➻ 2x + (x + 2 + 4) + (x + 4 - 7) = 27

➻ 2x + x + x + 6 - 3 = 27

➻ 2x + 2x + 3 = 27

➻ 4x + 3 = 27

➻ 4x = 27 - 3

➻ 4x = 24

➻ x = 24/4

➻ x = 6

• First integer is 6.

And,

➻ Second integer = x + 2

• Put x = 6 in above equation we get,

➻ Second integer = 6 + 2

➻ Second integer = 8

• Second integer is 8.

Also,

➻ Third integer = x + 4

• Put x = 6 in above equation we get,

➻ Third integer = 6 + 4

➻ Third integer = 10

• Third integer is 10.

Answer :

• Hence, required integers is 6 , 8 & 10.

Verification :-

We know that,

➻ 2x + (x + 2 + 4) + (x + 4 - 7) = 27

• Put x = 6 in above equation we get,

➻ 2(6) + (6 + 2 + 4) + (6 + 4 - 7) = 27

➻ 12 + 12 + (10 - 7) = 27

➻ 24 + 3 = 27

➻ 27 = 27

➻ LHS = RHS

• Hence, verified.

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Answer 2

Answer:

Sum of twice the first integer four more than the second integer and seven less than the third integer is 27. ... Answer : The integers are 6, 8 and 10 respectively .

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