Incidentally you can type absolute value bars || using the
key just above the ENTER or RETURN key. It is the shifted \.
|3x+2|-|x-3| = 5
There are 4 cases to consider, which are all 4 ways to
place signs in front on the expressions between the
absolute value bars:
+(3x+2)+(x-3) = 5, +(3x+2)-(x-3) = 5, -(3x+2)+(x-3) = 5, -(3x+2)-(x-3) = 5
3x+2+x-3 = 5, 3x+2-x+3 = 5, -3x-2+x-3 = 5, -3x-2-x+3 = 5
4x-1 = 5, 2x+5 = 5, -2x-5 = 5, -4x+1 = 5
4x = 6, 3x = 0, -2x = 10, -4x = 4
x = 6/4, x = 0 x = -5 x = -1
x = 3/2
We must check for extraneous solutions by substituting into the
original equation:
Checking x = 3/2
|3x+2|-|x-3| = 5
|3(3/2)+2|-|3/2-3| = 5
|9/2+4/2|-|3/2-6/2| = 5
|13/2|-|-3/2| = 5
(13/2)-(3/2) = 5
10/2 = 5
5 = 5
So 3/2 is a solution.
Checking x = 0
|3x+2|-|x-3| = 5
|3(0)+2|-|0-3| = 5
|0+2|-|-3| = 5
|2|-3 = 5
2-3 = 5
-1 = 5
So 0 is NOT a solution.
Checking x = -5
|3x+2|-|x-3| = 5
|3(-5)+2|-|-5-3| = 5
|-15+2|-|-8| = 5
|-13|-8 = 5
13-8 = 5
5 = 5
So -5 is a solution.
Checking x = -1
|3x+2|-|x-3| = 5
|3(-1)+2|-|-1-3| = 5
|-3+2|-|-4| = 5
|-1|-4 = 5
1-4 = 5
-3 = 5
So -1 is NOT a solution
The only solutions are 3/2 and -5
Edwin
