Question

Solve the equation |2x-1| - |x+5| = 3. Be sure to identify all possible cases and check your results.

Answer 1

Hey c:
Welcome to OpenStudy.

Hmm this one isn't too difficult...
It's just difficult to explain...
lemme think... hmm

Answer 2

The absolute value switches from positive to negative when the inside is 0.
So if you take absolute x as an example,\[\large\rm |x|\]It's positive x when we're larger than 0,\[\large\rm |x|=x,\qquad x\ge0\]and it's negative when we're less than 0,\[\large\rm |x|=-x,\qquad x<0\]

Answer 3

So how bout our two absolute values that we have here?
When do they switch from positive to negative?
When is the inside of the absolute 0?\[\large\rm 2x-1=0\qquad\to\qquad x=?\]\[\large\rm ~~x+5=0\qquad\to\qquad x=?\]

Answer 4

I agree. You should end up with two solutions, one each for each of these two equations. Once you have these two solutions, graph them on the number line. This will divide the number line into 3 parts. Choose a test number from each subinterval (of which there are 3) and determine whether the given absolute value equation is true or not. Repeat for the other subintervals. Write your solution in interval notation if appropriate.

Answer 5

Thank you for your help!

Answer 6

Was the question changed? Neither of the answers seem to make a lot of sense for the equation |2x-1| - |x+5| = 3

Answer 7

No...?

Answer 8

The person just said thank you...
so I didn't go into more detail :o

Answer 9

But @zepdrix why would you need to find where the absolute value part is equal to zero?

Isn't the way to solve these to first isolate one abs. value, then drop the signs, then repeat?

Answer 10

|2x-1| = 3+|x+5|
then
2x-1 = 3+|x+5| or 2x-1 = -(3+|x+5|)
then you isolate again, and solve.

Then you have to repeat the whole process, for the |x+5| this time.

Answer 11

I'm sure there are different ways to solve these but I guess I was approaching it like this:

The sign changes occur at \(\large\rm x=\frac12\) and \(\large\rm x=-5\)

So when \(\large\rm x<-5\), both absolutes are negative,\[\large\rm -(2x-1)--(x+5)=3\]Which leads to an extraneous solution.

When \(\large\rm -5\le x<\frac12\), the first absolute is negative,\[\large\rm -(2x-1)-(x+5)=3\]Leading to a solution.

And then there is the third region, \(\large\rm x\ge\frac12\).

Answer 12

I think you might lose information using only the plus/minus approach.
Like you might get more solutions than you are supposed to.
Hmm

Answer 13

Ah yeah I see. I've never done that way.

But you just test the solutions, as you would with any abs. value equation anyway.

Answer 14

Oh true true true :D

Answer 15

I've never considered your way. Makes sense, solves less equations I guess. With my way you have to solve... eight equations total, i think.

Answer 16

Might as well see how annoying my way really is (i've used it in the past)

|2x-1| = 3+|x+5|
then
2x-1 = 3+|x+5| or 2x-1 = -(3+|x+5|)

2x-1-3= |x+5| or...

ah screw it, I don't feel like it.