??Find three consecutive even integers whose sum is 180. What is the first (lowest) of these integers?

Question:how do u solve this?theres websites that say how to do it but i still dont understand

Answers:
(the short way)
180/3 = 60
so
60 + 60 + 60 = 180
then take 2 from the first and add 2 to the third gives
58 + 60 + 62 = 180

(the long way)
algebraically you need to find x,y,z such that
x + y + z = 180
since they are consecutive evens
y = x+2
and
z = y+2
= (x+2) +2 = x+4
so
180 = x+y+z = x+(x+2)+(x+4) = 3x +6
solving gives
x=58
so
y = x+2 = 60
and
z = x+4 = 62

∴ 58 + 60 + 62 = 180 as above
.
Consider the minimum even number is x. then the next even number will be x+2 and the third will be x+4. As per your question, all these three should add upto 180. So, write the following statement

x(min. number)+(x+2)+(x+4)=180
i.e. 3x+6=180
3x=180-6=174
x=174/3=58.
Now, x is the min. even number which is 58, so x+2 will be 60 and x+4 will be 62.
Bible and VD did it way more complicated than I did. I figured it had to group around 60 and came up with
58, 60 and 62.

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