Algebra Know-It-ALL

596 Worked-Out Solutions to Exercises: Chapters 1 to 9

4. It makes sense to consider warming as a positive temperature change, and cooling as a
   negative temperature change (although there is no technical reason why you couldn’t
   do it the other way around). Here are the year-by-year records once again
   for reference:
   
   
   • January 2005 averaged 5 degrees cooler than January 2004.
   
   
   • January 2004 averaged 2 degrees warmer than January 2003.
   
   
   • January 2003 averaged 1 degree cooler than January 2002.
   
   
   • January 2002 averaged 7 degrees warmer than January 2001.
   
   
   • January 2001 averaged the same temperature as January 2000.
   
   
   • January 2000 averaged 6 degrees cooler than January 1999.
   
   
   • January 1999 averaged 3 degrees warmer than January 1998.
   
   
   
   

If positive change represents warming, then going backward in time you have year-to-

year changes of −5, 2, −1, 7, 0, −6, and 3 degrees. You add these integers to discover the

change in average temperature between January 1998 and January 2005:

(−5)+ 2 + (−1)+ 7 + 0 + (−6)+ 3 = 0

In Hoodopolis, there was no difference in the average temperature for January 2005 as

compared with January 1998.

5. This works whenever b=a. You can then substitute a for b and get

a−a=a−a

You could also substitute b for a, getting

b−b=b−b

Obviously, either of these equations will hold true for all possible integers.

6. This always works if c= 0. With that restriction, a and b can be any integers you want.
   In order to see why, you can start with the original equation:

(a−b)+c=a− (b+c)

“Plug in” 0 for c. Then you get

(a−b)+ 0 =a− (b+ 0)

which simplifies to

a−b=a−b

That equation holds true for any a and b you could possibly choose. The values of a and

b don’t have to be the same.