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Step-by-step explanation:

Let x be the first integer

Let y be the 2nd integer

Let z be the 3rd integer

Given

[tex]x + y + z = 3y[/tex]

Also given, x = 7, y = 8, z = 9, substitute x,y and z into equation.

[tex]x + y + z = 7 + 8 + 9 \\ = 24[/tex]

[tex]3y = 3 \times 8 \\ = 24[/tex]

Therefore,

x+y+z = 3y for x = 7, y = 8, z = 9.

Now lets take x = 3, y = 4, z = 5.

[tex]x + y + z = 3 + 4 + 5 \\ = 12[/tex]

[tex]3y = 3 \times 4 \\ = 12[/tex]

As you can see , x+y+z = 3y is true for any 3 consecutive numbers.

ReimiSugimoto

Answer: (n-1) + n + (n+1) = 3n

Step-by-step explanation: So, for this equation, plug in the integers 7, 8 and 9, preferably in order. The variable next to the 3 will be the second integer. (7-1) + 8 + (9+1) = 3(8). If you solve the equation, you will get 24=24, which is true. This is because since the integers are consecutive, you start out with like 24=24. Then you subtract from the left side, which becomes 23, and then you add 1 again to it, which becomes 24. Hopefully, you understand what I mean. It's a difficult concept to explain.

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