Find three consecutive even integers such that the sum of the smallest number and twice the middle number is more than the largest number.

Hey there!!

Let's take the first number as 'x'

Then, the second consecutive integer would be 'x+2'

Hence, the third consecutive integer would be 'x+4'

As the question states, the sum of the smallest integer and the second integer is 20 greater than the third number.

Hence, the equation would be:

... x + x + 2 = x + 4 + 20

... 2x + 2 = x + 24

Subtracting x on both sides:

... 2x - x + 2 = 24

... x + 2 = 24

Subtracting 2 on both sides:

... x = 24 - 2

... x = 22

Hence, the smallest number is 22.

The second number is = 22 + 2

... 24

The third number is = 22 + 4

... 26

Hence, the numbers are 22,24 and 26.

Hope my answer helps!

{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