EgmatQuantExpert wrote:
Different methods to solve absolute value equations and inequalities- Exercise Question #5If ||x+3| - 12| < 13, what is the range of x?
Options a) (-4, 22)
b) (-28, -2)
c) (-28, 22)
d) (-28, 2) U (2, 22)
e) (-28, -2) U (4, 22)
dave13 - whenever I see "nested" absolute value inequalities,
I break the problem into pieces or stages.
That process often includes considering negative and positive cases.
I think that this approach is easier and more basic than the theoretical references I see.
Essentially, we simplify one part of the LHS for part of the time that we solve.
We have to find the range of solutions for a "nested" absolute value inequality.
That situation is complicated. We want to make it simpler.
1) Make the LHS simpler.Change the inside absolute value expression with a
placeholder for a little while.
I will let
\(u\) stand for \(|x+3|\)
\(|x+3|\) becomes \(u\). That is, \(u=|x+3|\)
I will "put back" \(|x+3|\) later.
Change the original: \(||x+3| - 12| < 13\)
Changed: NOW we have: \(|u - 12| < 13\)I use \(u\)
without brackets to solve more easily
BUT I do keep in mind that the brackets
are "in the background"
\(u\)
stands for an absolute value expression, |x + 3|
\(u\) is "really" an absolute value expression
One way to think about absolute value: it measures distance from 0.
By definition, distances cannot be negative.
Absolute value must be positive or 0.
Absolute value can never be negative.
3) Using \(|u - 12| < 13\) proceed as usual
Case 1: the value inside
\(|u-12|\) is positive. (LHS is positive)
Simply remove the absolute value bars
\(u-12<13\)
\(u<25\)That result is okay. RHS is positive.
Case 2:
\(|u-12|\) is negative*
(LHS
is negative. I used a shortcut to get from
\(-(u-12)<25\)
to \(u-12>-25\) . The shortcut is in the footnote*)
\(u-12>-25\)\(u>-13\)No. Cannot use this case. RHS is negative.
\(u\) is actually an absolute value expression (is really |x+3|).
Absolute values cannot be negative.
Ignore that case. Toss it in the trash.
Case 1 works. Case 2 does not work. Next stage.
5) Use
\(u<25\) from Case 1. "Put back" \(|x+3|\). Proceed as usual
We have yet another inequality:
\(|x+3|<25\) Case 1: LHS is positive
\(x+3<25\)
\(x<22\)Case 2: LHS is negative
\(x+3>-25\)
\(x>-28\), which is the same as
\(-28<x\)Those two values give us our range of solutions
\(-28<x<22\)
End values that specify this range are
\((-28,22)\)
I hope that helps.
Answer C
*Finding the negative case in an absolute value inequality. I took a shortcut.
Use |x + 3| < 25, from #5, above. The case in which the value inside the brackets is negative is written
x + 3 > -25
That inequality does not appear to be "LHS is negative." It is. I used a shortcut / mechanical rule.
Mechanical Rule - In an absolute value inequality, to find Case 2 in which LHS is negative:
Remove absolute value brackets, add a negative sign to RHS, and flip the sign.
Start: |x + 3| < 25 , negative case
• EITHER
- (x + 3) < 25
(-1) (x + 3) < 25
Divide by negative 1. Dividing an inequality by a negative number flips the sign.\((\frac{-1)(x+3)}{(-1)}> \frac{25}{-1}\)x + 3 > -25
Then simplify to x > -28
• OR
- (x + 3) < 25
-x - 3 < 25
-x < 28
(-1) * (x) < 28\(\frac{(-1)(x)}{(-1)}>\frac{28}{-1}\)x > -28 **Regarding the theoretical references (I am not sure they're helpful, but . . .)
In the statement "if x is less than 0, then |x| = -x "
The right hand side is NOT a negative number. It is the negative of a negative number: RHS is positive.
Try ANY negative number:
|-3| = 3
3 = -(-3)
The absolute value of -3 is 3, and that absolute value, 3, is the opposite of negative 3.
The negative sign in variable x "hides."
You may want to take a look at Bunuel 's post starting HERE. My post follows: Approaches: If x is negative, then |x| = -x . Finally, just below that post Bunuel here: finishes the discussion. _________________