Find 4 consecutive even integers such that when three times the smallest of the intagers is added to twice the largest their sum is 293

If x is the first of the integers then the statement is:

(x) , (x+2) , (x+4) , (x+6)

This means that the smallest is x and the largest is x + 6

so:
3(x) + 2(x+6) = 293
3x + 2x + 12 = 293
5x = 281
x = 56.2
This gives us a none integer (decimal)
What now?

Wait remember how x started as the lowest? what if x was the highest instead?

x, x-2, x-4, x-6
so:

3(x-6) + 2(x) = 293
5x = 315
x = 62.2

This is as close as have gotten to the answer
Cant seem to get an integer.
Maybe error in the question or some bad math on my part.

Value is 6 i think k

Well 96 divided by 6 is don’t know what you want when you say distributive property of multiplication