Faster way to solve absolute value problems.

[vittalk_usa]

Hi Ron,
I am looking for a better approach to solve absolute problems having 2 absolute expressions.
The method I have very long as follows. (pls pardon me for the looong email). Is there a faster way to solve it?
Can any problem having 2 absolutes like the one below be reduced to 2 cases
(x -3) = (2x -3) and (x-3) = -(2x -3), as mentioned in Eqn and VICs book?

critical point approach. questions/choices with only TWO ABS values : eg.
|x - 3| = |2x - 3|

a) Find critical points: ie.. the points at which the abs value becomes 0. Ignore the absolute signs for a moment and equate the expr within the abs sign to 0 and find the critical points. eg. x - 3 = 0 => x =3; 2x -3 =0 => x = 3/2. So the critical points are 3 and 3/2.

b) Draw a number line and mark these 2 points to get 3 regions
Eg. Here there will be 3 regions. X < 3/2, 3/2 <x< 3, x>3

c) For each of the 3 regions, do the following:

d) Step 1: Find the Sign Multiplier: To do this choose a number that satisfies the region (ie if the region is x > 3, choose say 4 or 5 etc..). Now plug this value, into both the abs expressions treated without the abs sign, ie |x -3| will become (x -3). Plugging in 4 => (4-3) = +ve . Similarly plugging 4 => (2*4-3) = +ve. So for the region x >3, both abs expressions are +ve.

e) Step 2: Solve the 2 sides using the sign info:
Ie. for the region x > 3, it is (x-3) = 2x -3 => x = 0 ......
Step 3: Ignore soln that contradict the region boudaries
This value of x should be within the boundaries of the
region (ie. in this case the region was x > 3 and we got
value of x as 0. This contradicts our assumption. So throw
the soln and the region and move on to next region.
Step 4: If we get the same unique value of x for all regions,
then we have found a solution. If we get multiple solutions,
then that statement is insufficient by itself. Try combining
to see if there is a common solution.


Full example:
Here critical points are 3 and 3/2.
Draw a number line and mark these 2 points. there will be 3 regions.
x< 3/2, 3/2<x<3, x>3

a) when x< 3/2
Both |x-3| and |2x-3| will be negative
so, we get, -(x-3) = -(2x-3) or x=0
which is a valid solution since we took x<3/2.

b) Now for 3/2<x<3
Find the sign multiplier:
Choose x = 2 (some number between 1.5 and 3)
Calculate the signs of each abs expression: (x -3) => (2-3) => -ve.
So |x -3| gets reduced to -(x-3)
Now apply the same x =2 on the next abs expr.
(2x -3) => (2*2 -3) => +ve
Multiply with the sign and reduce the double abs expr to plain expr.
So |x -3| = |2x -3| gets simplified to
(x -3) = -(2x -3)
Solve this, we get x=2
which is also correct since 3/2<2<3

c) Now for the last region, x>3
Both |x-3| and |2x-3| will be positive
x-3=2x-3
x=0 which is not possible since we have took x>3.
hence, the original problem is insufficient coz it gives two values of x. 0 and 2


Thank you very much.

[Ben Ku]

Using a sign chart is not incorrect, but it does take a lot of time and doesn't really help get to the answer. You will find that any double absolute value equation can be solved with the two possible equations below.
(x -3) = (2x -3) and (x-3) = -(2x -3)