Five times the first of three consecutive even integers is 2 less than twice the sum of the second and third. find the 3 integers

As we are being asked to solve for the middle of three consecutive even integers, let this be x.  Then, the first even integer is x - 2, and the third even integer is x + 2.

Then, we are given that 5 times the second of three consecutive even integers is six more than twice the sum of the first and third integers.

Translating this into an equation, we have

5 * x = 6 + 2 * (x-2 + x+2)

5x = 6 + 2 * 2x

5x = 6 + 4x

5x - 4x = 6

1x = 6

x = 6

Check the answer; the middle even integer is 6, so the first even integer is 4 and the third is 8

Then, 5 times the second even integer is 5 * 6 = 30

6 + 2 * the sum of the first and third integers is

6 + 2 *(4 + 8) = 6 + 2 * 12 = 30

30 = 30, so it checks

this doesn't make sense lol

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8