Hi, there--

The Problem: 
Find two consecutive even integers such that twice the smaller diminished by twenty is 
equal to the larger.

A Solution:
Let n be the smaller even integer.
The other even integer is two more than the first one, so the second integer can written 
as n+2.

In algebra, we write "twice the smaller diminished by twenty" as 2n-20.

An equation representing the relationship in  this problem is
n+2 = 2n-20

Solve for n to find the smaller integer. Subtract 2 from both sides.
n = 2n-22

Subtract 2n from both sides.
-n = -22

Divide both sides by -1.
n = 22

In the context of this problem n=22 means that the smaller even integer is 22. 
The larger even integer is two more, or 24.

Check your work by  going back to the original words of the problem.
Twice the smaller integer is 2*22= 44. "Diminishing" 44 by 20 gives 24. Check!

That's it. Please let me know if this is unclear, or if you still have questions,

Mrs.Figgy
math.in.the.vortex@gmail.com