Find four consecutive odd integers such that the sum of the second 
and the fourth is 63 more than one-fifth of the third.....that's my
question...can you show the solution?????

"Four consecutive odd integers" refers to the set {1,3,5,7} or
the set {3,5,7,9} or the set {9,11,13,15} or the set {77,79,81,83} or
any set of four such odd numbers in a row.  

To work such problems you start out by

1. letting x represent the smallest of four consecutive odd integers.

2. Then x+2 represents the second of four consecutive odd integers,

3. Then x+4 represents the third of four consecutive odd integers.

4. Then x+6 represents the fourth of four consecutive odd integers

>>...the sum of the second and the fourth is 63 more than one-fifth 
of the third...<<

Replace the words "the sum of the second and the fourth" by
(x+2) + (x+6)

So we have

>>...(x+2) + (x+6) is 63 more than one-fifth of the third...<<
  
Replace the words "one-fifth of the third" by "(1/5)(x+4)"  

So we now have

>>...(x+2) + (x+6) is 63 more than (1/5)(x+4)...<< 

To get 63 more than (1/5)(x+4) we add 63 to it, so replace
the words "63 more than (1/5)(x+4)" by "(1/5)(x+4) + 63" 

So now we have

>>...(x+2) + (x+6) is (1/5)(x+4) + 63...<< 

Finally replace the word "is" by "=" and we have completely
gotten rid of the English, and all that's left is algebra:

(x+2) + (x+6) = (1/5)(x+4) + 63

To solve that first multiply every term by LCD = 5. That is,
put 5 as a multiplier in front of every term on both sides

5(x+2) + 5(x+6) = 5(1/5)(x+4) + 5(63)

Replace the 5(1/5) on the right by 1, and the 5(63) by 315

5(x+2) + 5(x+6) = 1(x+4) + 315

Now distribute to remove parentheses, combine like signs,
solve, and you get

x = 31

So since the four cnsecutive integers are x, x+2, x+4 and x+6,
are respectively 31, 33, 35 and 37

Edwin