Solving Absolute Value Equations: Understanding Restrictions and Changing Values

[Nelo]

Urgent! Absolute value help

Homework Statement

Heres an example :: |5x-3| = |x+1|
we were given 3 cases for this type of problem in class they went something like this::

c.1 ) Use the 0 value for the right side and change all values in respect to (introduce - to all values) (x < -1)
c.2 ) -1< x < 3/5 , use the between and only change all leftside values to (-) keep R.S the same tho.

c.3) don't change anything and solve. x> 3/5 .

This is valid, however when I am doing other questions these "rules" don't seem to apply... for example :

|2-3x| = |5+x| I tried following the same outline, I got the correct answers, but none of them matched the restriction statement, they all went against the statement.

However, this method worked...

Start with the R.S ::: say x< -5 , and only change the L.S values
c2 ) use the between and change all values
c3) use x > 2/3 (other side) and only change right side values.

When i did this, the statements made sense corresponding with the restrictions. Why is this? Is there a piece of information that I am missing? How do i know which values to change and when to do so.

Homework Equations

The Attempt at a Solution

 

[Nelo]

anyone?

 

[Rayquesto]

Its hard for me to absorb the info you gave, but that's my problem and I'll just tell you what I think. I think that if there is an absolute value equation, the answer on the other side of the equation will be either positive or negative, meaning that there will be two answers opposite in direction, but since on either side of the equation, there contains an absolute value, the absolute value can cancel out I think.

so no matter what x=1

 

[Nelo]

Not the answer I am looking for.

Let me put it this way, How do you solve an equation with two absolute values on each side . Ie)

|5x-3| = |x+1|

|2-3x| = |5+x|

Do you have to use a different method to solve these? as in Two different methods, unique to each problem

 

[Mark44]

nelo,
It's much simpler than what you laid out.
If |5x-3| = |x+1|
then either
1) 5x - 3 = x + 1, OR
2) 5x - 3 = -(x + 1)

Solve both equations.

Rayquesto, there are two solutions.

 

[Nelo]

"sigh" , yes but there are 3 cases that need to be shown.

 

[Rayquesto]

thanks. I'm still a bit choppy on algebra skills, but most of the time these days, I don't need to be this specific to answer the more important complicated questions taken from calculus and physics, but it's always nice to know. So, thanks.

 

[Nelo]

lol k

 

[vela]

for example :

|2-3x| = |5+x| I tried following the same outline, I got the correct answers, but none of them matched the restriction statement, they all went against the statement.

However, this method worked...

Start with the R.S ::: say x< -5 , and only change the L.S values
c2 ) use the between and change all values
c3) use x > 2/3 (other side) and only change right side values.

When i did this, the statements made sense corresponding with the restrictions. Why is this? Is there a piece of information that I'm missing? How do i know which values to change and when to do so.

It's the negative coefficient of x on the LHS that's messing you up.

The function y=2-3x crosses the x-axis at x=2/3. When x<2/3, y>0, and when x>2/3, y<0. Similarly, y=5+x crosses the x-axis at x=-5. When x<-5, y<0, and when x>-5, y>0. The figure below summarizes this information.

       -6     -5     -4     -3     -2     -1      0      1      2
     <--+------+------+------+------+------+------+------+------+-->

2-3x ++++++++++++++++++++++++++++++++++++++++++++++++++o------------
 5+x ----------o++++++++++++++++++++++++++++++++++++++++++++++++++++

Because the coefficient of x in 2-3x is negative, the function is positive to the left of the zero and negative to the right of the zero. That's the opposite of what you get if the coefficient were positive, like with 5+x.The three cases are then:

Case 1: x < -5

2-3x is positive, so |2-3x| = 2-3x
5+x is negative, so |5+x| = -(5+x)

So you need to solve 2-3x = -(5+x).
​

Case 2: -5 < x < 2/3

2-3x is positive, so |2-3x| = 2-3x
5+x is positive, so |5+x| = 5+x

So you need to solve 2-3x = 5+x.
​

Case 3: x > 2/3

2-3x is negative, so |2-3x| = -(2-3x)
5+x is positive, so |5+x| = 5+x

So you need to solve -(2-3x) = 5+x.​

You might notice that cases 1 and 3 give you essentially the same equation. The first equation is just the second one multiplied by -1. So really, you're only solving the two equations

2-3x = 5+x
2-3x = -(5+x)

If you think about it, you should recognize it will always work out this way, which is why the faster method Mark gave above works.

 

[256bits]

"sigh" , yes but there are 3 cases that need to be shown.

In all fairness there are 4 cases here:
If |5x-3| = |x+1|
then either
1) 5x - 3 = x + 1, OR
2) -(5x - 3) = -(x + 1), OR
3) -(5x - 3) = x + 1, OR
4) 5x - 3 = -(x + 1)

You will notice that cases 1 and 2 are the same, and 3 and 4 are the same.
And that is where Mark44 simplifies to only 2 cases as he stated.
1) 5x - 3 = x + 1, OR
2) 5x - 3 = -(x + 1)

 

[Mark44]

The idea underlying what I did is this:
If |A| = |B|, where A and B represent some expressions,
then A and B are the same distance away from 0.

If A and B are the same sign, either both positive or both negative, then A = B.
If A and B are opposite signs, then A = -B.

In summary, |A| = |B| \Rightarrow A = B or A = -B