Question

the sum of an integer and twice the next integer is 41. find the two integers with method how 14 ans cmae in this

Answer 1

Let the required two consecutive integers are a and a + 1,

According to the question :

= > Sum of integer + twice of next integer = 41

= > a + twice of ( a + 1 ) = 41

= > a + 2( a + 1 ) = 41

= > a + 2a + 2 = 41

= > 3a = 41 - 2

= > 3a = 39

= > a = 39 / 3

= > a = 13


Hence,
Required numbers are a = 13 and a + 1 = 13 + 1 = 14


Thus,
Required two integers are 13 and 14.
$\:$

Answer 2

• Integers :

  The set which is made with the help of whole numbers, like 0, ± 1, ± 2, ± 3, ... is called the set of Integers. The set of natural numbers, denoted by ℕ is the set of all positive integers ℤ⁺.

• Solution :

Let us consider the two consecutive integers to be x and (x + 1)

By the given condition, the sum of the first integer and twice the next integer is 41

  ⇒ x + 2 (x + 1) = 41

  ⇒ x + 2x + 2 = 41

  ⇒ 3x = 41 - 2

  ⇒ 3x = 39

  ⇒ x = $\dfrac{39}{3}$

  ⇒ x = $\dfrac{\cancel{3}*13}{\cancel{3}}$

  ⇒ x = 13

∴ The two consecutive integers are

  13 and (13 + 1)

  i.e., 13 and 14

• Verification :

  The 1st integer = 13

  Next integer = 14

∴ 1st integer + twice the next integer

  = 13 + 2 ( 14 )

  = 13 + 28

  = 41

Thus, verified.