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

is there a picture

step-by-step explanation:

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1