Find three consecutive even integers so that the first plus twice the second is twice the third

Answer 1

1st # -> x

2nd # -> x + 2

3rd # -> x + 4

first # + twice the 2nd# = twice the 3rd

x + 2(x + 2) = 2(x + 4)

Now simplify and solve the resulting equation.

distribute on both sides x + 2x + 4 = 2x + 8

combine like terms on the left side 3x + 4 = 2x + 8

move 2x from the right to the left side -2x -2x

simplify x + 4 = 8

now undo the add 4 by subtracting 4 -4 -4

simplify and you have your answer for x x = 4

Therefore your 1st # is 4, the 2nd # x + 2 = 4 + 2 = 6, the 3rd # x + 4 = 4 + 4 = 8