Solving equation containing absolute values

AI Thread Summary
To solve the equation |2x+7| - |6-3x| = 8, it is important to consider three cases based on the sign of the expressions within the absolute values. First, determine where each expression is positive or negative: 2x+7 is positive for x > -7/2, and 6-3x is positive for x < 2. The three cases to analyze are when both expressions are positive, when one is positive and the other is negative, and when both are negative. By applying this method, you can find all possible solutions, ensuring to check the validity of each solution based on the assumed conditions. This structured approach will help in accurately solving the equation.
jkristia
Messages
54
Reaction score
0

Homework Statement



I have this equation

|2x+7| - |6-3x| = 8.

The step I did is to replace the || with () and then solve the equation
2x+7-6+3x = 8
X = 7/5

But how do a go about solving for the second solution?
With one absolute value I would
|2x + 7| = 8
2x + 7 = +-8
2x = -7 +-8
X = -15/2, x = ½

But I can’t see how to solve the first equation for the second solution.
 
Physics news on Phys.org
The best way to approach this type of problem is to take 3 cases.

Firstly, you need to find at which x value each absolute value is positive or negative. 2x+7>0, then x>-7/2, thus for x>-7/2 that expression is positive and for x<-7/2 that expression is negative.
6-3x>0, x<2, thus for x<2 it is positive, and x>2 it is negative.

Now the first case you should consider is for all x values such that both expressions are positive.
The next case should be for x values where one is positive and the other is negative.
The last case should be when both are negative.

As an example, if we have to solve |x|+|x-1|=2, for x>1 both are positive, so we simply solve x+(x-1)=2, for 0<x<1 we have the first being positive and the second being negative, thus we solve x-(x-1)=2 and for x<0 both are negative so we solve -x-(x-1)=2.

Just apply the same idea to your question.

p.s. remember that since we assumed x<0 for the last case, the solution needs to be less than zero, else it is an invalid solution and you just discard it.
 
I think I got it, thank you for your help.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top