Solving an Absolute Value Inequality with Two Terms

[projection]

Homework Statement

[tex]\left|x-5\right|\leq\left|x+3\right|[/tex]

i know how to slove abs inequalities when there is one. like [tex]\left|x-5\right|\leq5[/tex]. put x-5 between 5 and -5. then solve for x.

i don't know how to start this.

can someone provide some insight.

 

[Dr Zoidburg]

Homework Statement

[tex]\left|x-5\right|\leq\left|x+3\right|[/tex]

i know how to slove abs inequalities when there is one. like [tex]\left|x-5\right|\leq5[/tex]. put x-5 between 5 and -5. then solve for x.

i don't know how to start this.

can someone provide some insight.

pretty rusty myself with abs inequalities, but can you just do the same as what you mentioned [tex]\left|x-5\right|\leq5[/tex].
i.e:
[tex]\left|x-5\right|\leq\left|x+3\right|[/tex]
=

[tex]x-5 \leq\ x+3 \leq\ x+5[/tex]
and then solve from there?

as I said, I'm rusty on these things so apologies if I'm leading you up the garden path!

 

[HallsofIvy]

A method I prefer is to solve the associated equation first!
If |x- 5|= |x+ 3| then either x- 5= x+ 3 or x- 5= -(x+ 3). The first reduces to the equation -5= 3 which is NEVER true and so does not give a solution to the equation. The second is the same as x- 5= -x- 3 or, adding x and 5 to both sides, 2x= 2 so x= 1

Now, the point is that x= 1, where |x-5|= |x+ 3| separates |x-5|< |x+3| and |x- 5|> |x+ 3|. check one value of x on each side of 1 (x= 0 and x= 2 are simple) to see on which side |x-5|< |x+3|.

You might also remember that |x- a| can be interpreted as "distance from x to a". |x- 5| can be interpreted as "distance from x to 5" and |x+3|= |x-(-3)| as "distance from x to -3". Asking "where is |x-5|< |x+3|?" is the same as asking "what points on the line are closer to 5 than to -3?". You might notice that, geometrically, x= 1 is exactly half way between 5 and -3.

 

[Tedjn]

Just so you get exposed to this, another way to think about the problem is follows:

[tex]|x - 5| \leq |x + 3| \implies (x-5)^2 \leq (x+3)^2 \implies x^2-10x+25 \leq x^2+6x+9 \implies 16 \leq 16x \implies 1 \leq x[/tex]

Notice that the x² cancels out so we only have one solution.

 

[Dr Zoidburg]

A little off-topic (and this has prob been well-documented already, but apologies: I am new here!), but I wrote:
[tex]x-5\leq\x+3\leq\x-5[/tex]
then realized the error and edited it to:
[tex]x-5\leq\x+3\leq\x+5[/tex]
but it keeps showing up as the original on screen. If you click on it for the LaTeX code reference, it shows it as the corrected version.


why is this?

edit: now the x is missing! what am I doing wrong here?

 

[Tedjn]

That is because in your most recent LaTeX, you have a backslash before the x, which shouldn't be there. For your other problem, make sure you erase the browser cache when you refresh. Most times, on Windows, it is Shift+F5 or Ctrl+F5 instead of just F5.

 

[Dr Zoidburg]

cheers. wanted to clear that up before I make any more screw-ups.

 

[projection]

Just so you get exposed to this, another way to think about the problem is follows:

[tex]|x - 5| \leq |x + 3| \implies (x-5)^2 \leq (x+3)^2 \implies x^2-10x+25 \leq x^2+6x+9 \implies 16 \leq 16x \implies 1 \leq x[/tex]

Notice that the x² cancels out so we only have one solution.

is there an explanation as to why this would work?

 

[Tedjn]

It works really because each statement follows from the previous one. If you want a more conceptual way to think about it, I think HallofIvy's idea about picturing absolute value as distance on a number line is the best choice. In any case, from that picture, basically this takes the square of that distance, and finds what values of x work.

 

[HallsofIvy]

It works because |x|²= x² for any real number, x.