Question

find three consecutive integers such that the sum of the first and the second is 6 more than the third

Answer 1

Hello!

Let the consecutive integers be x, x + 1 and x + 2

✓ The sum of 1ˢᵗ and twice the 2nⁿᵈ :

x + 2(x + 1) = 56 - 3(x + 2)

⇒ 3x + 2 = 56 - 3x - 6

⇒ 6x = 48 ⇒ x = 8

Hence,
The required integers are :

• x = 8

• x + 1 = 8 + 1 = 9

• x + 2 = 8 + 2 = 10

Cheers!

Answer 2

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