Find three consecutive even integers, such that 9 times of first integer is 8 more than the sum of second and third integers

Let the middle number = X

The other two numbers are even so the lowest number would be X -2, and the highest number would be X +2.

9 times the first number ( lowest number) would be 9(x-2) = 9x-18

The sum of the other two numbers becomes x + x+2 = 2x+2

The first number is 8 more so now you have an equation of :

9x-18 = 2x+2 +8

Combine like terms on the right side:

9x-18 = 2x+10

Add 18 to both sides:

9x = 2x + 28

Subtract 2x from each side:

7x = 28

Divide both sides by 7:

X = 28 / 7 = 4

The middle number is 4.

The lowest number = X-2 = 4-2 = 2

The highest number = X+2 = 4+2 = 6

The 3 numbers are 2,4 & 6

idk this answer sorry cant on this one.

step-by-step explanation:

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: