Example I: Find the solution set S of the
equation |x| =5. Answer: S = {+5, -5} or x =
(+/-) 5
Example II: Find the solution set S of the
equation |Q| =5. Answer: S = {+5, -5} or Q =
(+/-) 5 (the letter used in the equation does not matter)
Example III: Find the solution set S of the
equation |x-3 | = 13.
Answer: Put P = x -3. The equation says |P| = 13 and
hence the P = (+/-)13. So we know that
P = -13 or P = +13.
Since P is shorthand for x - 3, these two case imply respectively that
x -3 = -13 or x - 3 = 13
and hence x = -13 + 3 = -10 or x = 13 + 3 = 16.
So we suspect the solution set S = {-10, 16) or equivalently x = 3 +
(+/-)13. That completes the answer except for a check - checks are always
needed for multi-step answers or processes.
Check: When x = 16, x-3 = 13 and |x-3| = 13
and when x = -10, x-3 = -10 -3 = -13 and |x-3| = 13 again. So our answer
work.
Remark: Our values for x are the only answers as we followed chains
of reason which implied if |x-3| = 13 then x = -10 or x = 16.
Example IV: Find the solution set S of the
equation |x-3 | < 13.
Answer: Put P = x -3. The equation says |P| < 13 and
hence the P belongs to the interval (-13, 13) with end points included
(why?)
Since P is shorthand for x - 3, we require x - 3 belong to the interval -13
to 13 (endpoints excluded). That gives two inequalities to satisfy,
namely
x -3 > -13 and x - 3 < 13
The first requires x > -13+3 = -10 while the second requires x
< 13 + 3 =16.
So we must have -10 < x < 16. So the solution set S =
(-10, 16).
Check: If - 10 < x < 16, we may subtract 3 to find
-13 < x - 3 < 13 and hence |x-3| < 13.
Remark: The endpoints of the solution interval -10 < x < 16
would be included if we were solving |x-3 | < 13 instead of |x-3 |
< 13. Do you see why.
Example V: Find the solution set S of the equation
( | x+3 | - 4 )(x+8) = 0.
Answer: The only way ( | x+3 | - 4 )(x+8) = 0 is if at least
one of the factors is zero. So we must have
( | x+3 | - 4 )= 0 or (x+8) =0
The second factor x+8 = 0 when and only when x = -8. So our first solution
is x = -8.
The first factor is zero when and only when |x+3| = 4 or equivalently x+3 =
(+/-)4 or x = - 3+ (+/-)4. So the first factor is zero when and only
when x = -3+4 = 1 or x = -3 - 4 = -7.
So the first factor is zero for x = 1 or x = - 7 while the second
factor is zero for x = -8.
So in the absence of mistakes in our reasoning, any solution of the ( | x+3
| - 4 )(x+8) = 0 must be an element of the set S = {1, -7, 8}
Now complete the solution by checking that these values work.
Example VI: Find the solution set S of the
equation | x2-17 | = 8.
Solution: x2-17 = (+/-)8. Therefore x2
= (+/-)8. +17 belongs to { 17 - 8, 17+ 8} = {9, 25}. That is
x2 = 9 or x2 = 25.
So x = (+/-) 3 or x = (+/-)5 or equivalent the solution set S = { -3, +3,
-5, +5}
Now complete the solution by checking that these values work.