Question

Q7. Find three consecutive even integers such that the sum of twice th
first integer , four more than the second and seven less than the third
integer is 27.. Find the integers.​

Answer 1

Answer:

Given :The sum of the first, twice the second, and three times the third = 70

To find : The three consecutive odd integers.

Proof:

Let the first odd integer = x ,

The second consecutive odd integer = x+2 ,

The third consecutive odd integer = (x+2)+2 =  x+4

So,The sum of the first, twice the second, and three times the third will be

x+2(x+2)+3(x+4)

Given the sum of the first, twice the second, and three times the third = 70

∴   x+2(x+2)+3(x+4) = 70

=> x+2x+4+3x+12 = 70

=> 6x+16 = 70

=> 6x = 70-16

=> 6x = 54

=> x = 54÷6

=> x = 9

Now we have;

first odd integer = x

=> x= 9

The second consecutive odd integer = x+2

=> x+2

=> 9+2 = 11

=> x = 11

The third consecutive odd integer = x+4

=> x+4

=> 9+4 = 13

=> x = 13

∴ the three consecutive odd integers x=9, x=11, x= 13.

Hence, the three consecutive odd integers are 9,11,and 13.

Step-by-step explanation:

BRAINLES PLEASE