Question

Find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third. *

Answer 1

Answer:

6,8,10

Step-by-step explanation:

Let those numbers are a, a + 2, a + 4.

sum of the smallest integer and twice the second is 12 more than the third.

=> a + 2(a + 2) = (a + 4) + 12

=> a + 2a + 4 = a + 4 + 12

=> 3a + 4 = a + 4 + 12

=> 3a - a = 4 + 12 - 4

=> 2a = 12

=> a = 12/2 = 6

Numbers are,

a = 6 ; a+ 2 = 8 ; a + 4 = 10

Answer 2

Answer:

$\underline{\boxed{\red{\bf 6,8,10}}}$

Step-by-step explanation:

Let the three consecutive even integers be x, (x + 2) and (x + 4) respectively.

According to the question,

The sum of the smallest integer and twice the second integer is 12 more than third.

Here we go ;

⇒ x + 2(x + 2) = 12 + (x + 4)

⇒ x + 2x + 4 = 12 + x + 4

⇒ 3x + 4 = x + 16

⇒ 3x - x = 16 - 4

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$

$\boxed{\bf \therefore x = 6}$

Thus, the required numbers are :

• x = 6
• x + 2 = 6 + 2 = 8
• x + 4 = 6 + 4 = 10

Therefore, the three consecutive even integers are 6, 8 and 10.