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find three consecutive even integers such that the sum of the first and second equals 3 times the Third

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Find three consecutive even integers such that the sum of the first and second equals 3 times the third​.

We have to set up an equation that satisfies this statement.

Notice that consecutive even integers like 2, 4 and 6 , has a common difference of 2 ( such that 4-2 = 2 and 6-4 is also 2 and so on) . We can represent the three consecutive integers by letting:

x - 1st even integer

x + 2 - 2nd even integer

x + 4 - 3rd even integer

It is stated that the sum of the first two integers equals three times the third; the expression that describes this can be written s follows:

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

*sum of the first two integers - x + ( x + 2 )

*three times the third integer - 3 ( x + 4)

Solving for x

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

x + x + 2 = 3x + 12

2x + 2 = 3x + 12

3x - 2x +12 - 2 = 0

x + 10 = 0

x = -10

thus,

x + 2 = -10 + 2 = -8

x + 4 = -10 + 4 = -6

ANSWER:

-10, -8 and -6

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