## how to solve |4x|=|4x+1| Is this resolution right?

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rulolp
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### how to solve |4x|=|4x+1| Is this resolution right?

Hi, first thanks for your help, and for the page!, and sorry for my English, I'm from Argentina
I was looking the example about the absolutes, and I find |x–3|=|3x+2|–1 that's similar what I'm searching (and when I put resolve graphically the example in the page it plotted me another thing, I send a message asking for that).
I put the expression |4x|=|4x+1| in plotters online but tell me that's an invalid expression, and in the plotter of this page it show me another graphic too.
Hope you can help me, telling me if both resolutions are done (Isn't the first way more easy?), and the arguments are fine. At the end I ask about the absurds finded in the solution, I remain dobious about it.

When I solve the exercice I put
|4x|=|4x+1|
4x = + - |4x+1|
that give me 4x = 4x+1 , -4x=4x+1 (both in case +|4x+1| when I take the bars of)

And +/- (-4x) = 4x + 1 when the case is -|4x+1| by multiplying for -1 both sides, and taking the bars off. That give me 4x=4x+1 and -4x=4x+1 the same cases above

In the case of 4x=4x+1 -> 0=1 that's an absurd or ilegal expression don't?
In the case of -4x=4x+1 -> -8x=1 -> x=-1/8

That's the same solution if I do by the method explained here, looking the intervals
|4x| >= 0 for x>= 0 doing 4x >= |0| and then x >=0/4 then x>=0
|4x+1| >= 0 for x>= (-1/4)
I know Y=4X has a positive slope and Y=4X+1 too.
That's points, 0 and -1/4, are where the absolute-value expressions equal zero, giving me the intervals, (-infinity,-1/4),(-1/4,0),(0,infinity).
In the first interval, both absolute values are negatives, so I have -4x=-4x-1 -> 0=-1 that's illegal
the same thing happens in the last interval 0=1
but in the middle -4x=4x+1 then -8x=1 and then x=(-1/8) that's between -1/4 and zero and is a valid solution.
Then the answer is X=(-1/8), the same above.

That's fine? what's means I have an absurd? in the example in the page get an invalid solution because they have out of interval solution, but I get an absurd, that's fine? the solutions are the points of intersections of the lines, but the invalid solutions are nothing in the graphic it seems?

Thanks, and sorry for the extension!

maggiemagnet
Posts: 358
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### Re: how to solve |4x|=|4x+1| Is this resolution right?

Thank you for showing your working and thinking so nicely! Yes, these type equations sometimes come up with dumb answers, which means they come up with no answers.

There's a lesson here on solving. For yours we'll start by doing the intervals:

4x < 0 for x < 0
4x + 1 < 0 for 4x < -1 for x < -1/4

So we get (-infty, -1/4), [-1/4, 0), and [0, +infty). Now do on each:

(-infty, -1/4): 4x < 0, 4x + 1 < 0:

|4x| = |4x + 1|
-(4x) = -(4x + 1)
-4x = -4x - 1
0 = -1 (which is dumb)

So no solution.

[-1/4, 0): 4x < 0, 4x + 1 >= 0:

|4x| = |4x + 1|
-(4x) = 4x + 1
-4x = 4x + 1
-8x = 1
x = -1/8

This is in the interval so it's okay.

Now you do the last interval.

rulolp
Posts: 3
Joined: Tue Jan 07, 2014 4:41 pm
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### Re: how to solve |4x|=|4x+1| Is this resolution right?

Thanks! Yes, I saw the page you tell me http://www.purplemath.com/modules/solveabs.htm page 2, that have an example very similar. At the end I comment another answer about the topic and the method, but first:
The last interval I think would this way:
(0,infinity): 4x >= 0, 4x+1 >=0 so both expression must be positives.
|4x| = |4x+1|
4x = 4x+1
0=1 (which is dump)

Thanks for your answer, I understand now why think in intervals, but I don't understand why the other method doesn't work,
example:
|x-5|=4 taking the bars off ( x-5 if x-5>0 and 5-x if x-5<0 ) :
x-5 = +/- 4
so x= 9 or x=1

If I have
|4x| = |4x+1|
by the same argument previously
4x = +/- |4x+1|
having two ways, 4x = |4x+1| and 4x = - |4x+1|
going the first +/- 4x = 4x+1
4x=4x+1 which is dump, and -4x=4x+1 which is x = -1/8
and second option give me +/-(-4x)=4x+1 with same results.
That's seems to be OK, the solution is -1/8

But if I try with the example:
|x-3| = |3x+2| - 1
Following the same reasoning
x-3 = +/- ( |3x+2| -1 )
that give me two options:
A) x-3 = |3x+2| - 1 and B) x-3 = -( |3x+2| -1 )
going by the first option
A) x-2 = |3x+2|
+/- (x-2) = 3x+2 having
A1) x-2 = 3x+2
2x = -4
x = -4/2 = -2
A2) 2-x = 3x+2
4x = 0
x = 0

B) x-3 = -( |3x+2| -1 )
x-3 = -|3x+2| + 1
x-4 = -|3x+2|
4-x = |3x+2|
+/- (4-x) = 3x+2
B1) 4-x = 3x+2
4x = 2
x = 2/4 = 1/2
B2) x-4 = 3x+2
2x = -6
x = -6/2 = -3

The solutions B1 and B2 are right, but no A1 and A2, but the reasoning was the same, I'm wrong? Why I can't trust in taking the bars off method?
I thought the method of take the bars off was more simple, but it didn't give me a good answer in the last exercice.

Well, thanks for your time, thank you very much, really, I haven't anybody to question at the momment, we are in vacations and didn't find many examples. I hope this questions help other people like me
Greetings!

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
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### Re: how to solve |4x|=|4x+1| Is this resolution right?

If I have |4x| = |4x+1|...
...
That's seems to be OK, the solution is -1/8
One advantage of intervals is that you're sure the answer you get is right. Sometimes solutions look okay but aren't right for the interval, like "x = 3" when the interval is "x < 0". It's okay for what you did, but sometimes the interval is going to show the solution isn't right.
|x-3| = |3x+2| - 1
Following the same reasoning
x-3 = +/- ( |3x+2| -1 )
You're doing the +/- for the "x - 3", but what about the "3x + 2"? Because you've included the "- 1" with the absolute value, you're not getting the right answers.

rulolp
Posts: 3
Joined: Tue Jan 07, 2014 4:41 pm
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### Re: how to solve |4x|=|4x+1| Is this resolution right?

rulolp escribió:
|x-3| = |3x+2| - 1
Following the same reasoning
x-3 = +/- ( |3x+2| -1 )

You're doing the +/- for the "x - 3", but what about the "3x + 2"? Because you've included the "- 1" with the absolute value, you're not getting the right answers.
Sorry, I can't understand you, what's wrong?, to solve the "|3x+2|" first I need to move the "-1" to left side as positive or negative, if I choice the + road, the -1 keep equal and pass to the left side as positive, keeping x-2 left side and |3x+2| right side, now I can take off the bars, but this line give me the wrong results. Maybe I have a bad conversion somewhere, but can't see it. Only the - road give me the good responses. The -2 or 0 results should be dump to invalidate results like 0=1 as previously examples, or be 1/2 or -3 to be valid.

So far, I tried to see the range, if the expression was exp1<exp2 or remove bars if was exp1=exp2 it seems to be more simple or easy to solve. But in the three exercices showed, the first two give me good answers, but the third don't, can you give me the resolution taking off the bars please? I can't see the mistake.

Another topic is how to write the final solution, can be "S={(1/2,-3)}" and in the case of an open interval be "S=(-2,-1)" like |2x+3|<1 expression?

Greetings!, and sorry for my english if you find an error.