# Absolute value equation problem...can you check my work?

Discussion in 'Undergraduate Math' started by Will H, Apr 19, 2007.

1. ### Will HGuest

|1/2x - 1/3| = 1/5|-x|

Solve the equation.

For real numbers y and c such that c> 0.
|y| =c iff y = c OR y = -c

Therefore...

1/2x -1/3 = 1/5|-x| OR 1/2x -1/3 = -1/5|-x|
1/2x-1/5|-x|=1/3 1/2x+1/5|-x|=1/3
15/30x - 6/30|-x|=10/30 15/30x + 6/30|-x| = 10/30
9/30x = 10/30 21/30x =10/30
x=10/9 x= 10/21

Is this correct?

Will H, Apr 19, 2007

2. ### VirgilGuest

If you substitute 10/9 or 10/21 back in the original equation in place
of x, does it balance?

Virgil, Apr 19, 2007

3. ### Arturo MagidinGuest

I assume "1/2x" means "(1/2)x" and not 1/(2x), given what you write
later. Please try to be unambiguous.
You can do better here. See below.

Of course, this step in isolation is incorrect. You cannot go from the
line immediately previous to the bottom, because you have to consider
two cases again: whether x>=0 or x<0. As it happens, though, the four
cases you get (one from each column) only amount to the two cases you
describe.
Looks like. But you can do better:

If (1/2)x - (1/3) = (1/5)|-x|, then you know that (1/2)x-(1/3) is
positive. That means that (1/2)x - (1/3) > 0, so x>2/3. In particular,
|-x|=x, so you can immediately replace the equation with

(1/2)x - (1/3) = (1/5)x

and solve this equaltion, which gives

(1/2)x - (1/5)x = (1/3)

(3/10)x = (1/3)
x= 10/9.

Now, suppose that (1/2)x - (1/3) = -(1/5)|-x|. That means that
(1/2)x - (1/3) is negative. Hence

(1/2)x < (1/3)
x < (2/3).

Now, you cannot figure out from this whether |-x|=x or |-x|=-x.

If 0<x<2/3, then |-x|=x, and you get

(1/2)x - (1/3) = -(1/5) x
(1/2)x + (1/5)x = (1/3)
(7/10)x = (1/3)
x = 10/21.

On the other hand, if x<0, then |-x|=-x, so you get

(1/2)x - (1/3) = -(1/5)(-x)

however, this leads to x=10/9, which is incompatible with the
assumption that x<0, so this case cannot occur. So you are left with
only the two solutions you had.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Arturo Magidin, Apr 19, 2007
4. ### Arturo MagidinGuest

That would show Will that the answers he got were correct. But it
would not ensure him that he has managed to find ALL solutions.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Arturo Magidin, Apr 19, 2007
5. ### Greg NeillGuest

Rather than struggling with the absolute values,
why not recognize that

|x/2 - 1/3| = (1/5)|-x| assuming x is real

implies (x/2 - 1/3)^2 = (1/25)x^2

Solve in the usual manner for x.

Greg Neill, Apr 19, 2007
6. ### Mike TerryGuest

This follows from your statement that I've labeled ,
since 1/5|-x| > 0.

However, you've still got a mix of x and |x| in each column, so this is not
best way to proceed. See suggestion below...
This is wrong, because you have just replaced |-x| with x, which is only
valid if x<0. You could proceed by examining the separate cases x<0, x>=0,
so you would then have 4 columns to consider, but you could have avoided
this pickle with a better starting direction...
A better approach would be to try and eliminate all the absolute values in
one go. You can use either of the following:

For real numbers y and z
|y| = |z| iff y = z OR y = -z  or
|y| = |z| iff y^2 = z^2 

Your initial equation can very easily be put in this form, and using either
of  or  will set you off down a path where the equations you are
trying to solve do not have any remaining absolute value functions.

If you use  you will have two columns to follow through. (In fact this
will be in practice the steps you used above.) This will lead you to two
(possible) solutions by solving two linear equations.

Alternatively, using  you will have one column, but you will need to
solve a quadratic equation, which will again lead to two possible solutions.
Is one quadratic easier than two linear equations? (You decide! In either case, you will then only have succeeded in proving that a solution
to the original equation must be one of the two values you have obtained.
Next you will need to determine whether either of these two values *in fact*
satisfy the original equation. You do this simply by substituting the
values back into the original equation and checking if they work. (This
step was missing from your solution above!)

Regards,
Mike.

Mike Terry, Apr 19, 2007
7. ### Mike TerryGuest

Upon reflection, all the steps in the solution are reversible, due to the
"iff"s appearing in , , and . So solutions to the final equation
must also be solutions to the original equation, and there is no requirement
to confirm these solutions by substituting back into the original equation.

However, this is good practice in any case just to confirm you've not
slipped up somewhere with a silly mistake...

Mike Terry, Apr 19, 2007