Absolute Value - One Solution

[EricUva]

Can you please elaborate on the rules for testing multiple solutions on absolute value equations? Why does this question only have one solution? (This is a MGMAT problem)

Question:

What is x?

(1) |x| < 2

(2) |x| = 3x - 2

Specifically, why does statement 2 ONLY equal 1 (and not 1/2).



Thanks Again,

Eric

[Jeff]

Eric -

For the equation 3x-2=|x| , 1/2 doesn't satisfy the equation. Plug it in:

3*(1/2)-2=|1/2|

-1/2= 1/2 > Nope, doesn't work.

Here is how to methodically find the roots when dealing with an absolute value equation. You need to consider what happens when the value in the absolute value bars is both positive and negative.

First case: Value in absolute value bar is positive or zero

This means means that X>=0 so:

3x-2=x
x=1

To be a valid solution to the absolute value equation, the root must both satisfy the modified equation (3x-2=x) and be in the region of interest (X>=0). Since x=1 satisfies both requirements, it is a valid solution.

Second Case: Value in absolute value bar is negative

This means the area of interest is X<0

3x-2=-x
x=1/2

In this case 1/2 satisfies the modified equation (3x-2=-x) but is not in the area of interest (x<0)...meaning the expression within the absolute value bars wouldn't be negative. So 1/2 is not a valid solution.

What's happening with absolute value equations can be really tricking to grasp - graphing some of these equations can really be an aid in understanding what's going on and why the technique outlined above works.

Jeff

[EricUva]

Jeff, thanks for the quick response. Can you please critique my logic on the following MGMAT question?

|3 - y| = 11

According to the text, there are two solutions.

If y>=0, then y=-8.
If y<0, then y=14

To play devil's advocate here: Shouldn't the answer be "no solution"? Since when testing for y>=0, the answer is negative and when testing for y<0, then answer is positive. What am I missing?

thanks

Eric

[Jeff]

Eric -

|3-y|=11

This one is more straighforward than the previous example because it does not feature the variable both inside and outside the absolute value bars. So you don't actually need to wory about the regions over which the solution is valid...but here's how you'd do it anyway.

First consider the case where 3-y is positive:

3-y>0
-y>-3
y<3 : This is the valid region.

3-y=11
-y=8
y=-8 : good solution to the equation and in the region y<3

Now consider 3-y<0

3-y<0
-y<-3
y>3 : This is the valid region

3-y=-11
-y=-14
y=14: good solution to the equation and in the region y>3

Again, you only need to worry about being in the valid region when the variable appears both within and outside the absolute value bars. But if you have any doubt about your answer, you can always plug it into the original absolute value equation.

/Jeff

[Eric]

BTW, I believe the solution should read:

If (3-y)>=0, then y=-8.
If (3-y)<0, then y=14

Jeff

[jayant.apte]

Hi Jeff,

I have a question on the explanation you provided

For the equation 3x-2=|x| , 1/2 doesn't satisfy the equation. Plug it in:

3*(1/2)-2=|1/2|

-1/2= 1/2 > Nope, doesn't work.

and

In this case 1/2 satisfies the modified equation (3x-2=-x) but is not in the area of interest (x<0)...meaning the expression within the absolute value bars wouldn't be negative. So 1/2 is not a valid solution.

You used different approaches to check whether x=1/2 meets the requirements. First, by plugging it into the original equation 3x-2=|x| (including the absolute expression) and second, by comparing it against the region x < 0 (the region that satisfies the absolute expression) 3x-2 = -x.

So would it be correct to use either of these methods to ensure that a value of a variable (x=1/2) meets the equation requirements?

[RonPurewal]

the above posts are making my eyes hurt, because i'm lazy and don't like solutions with lots of steps.
also, there's no way that most people are going to be able to execute all those steps within the requisite time. (remember that this is just one statement, not even the whole problem!)

here's a takeaway that should help you ladies and gentlemen:

if you get
| QUANTITY1 | = QUANTITY2
or
| QUANTITY1 | = | QUANTITY2 |

then just
solve the following two equations:
QUANTITY1 = QUANTITY2
QUANTITY1 = -(QUANTITY2)

and then
CHECK your solutions, in the ORIGINAL EQUATION.


that's it, folks. no worrying about "areas of interest" or signs in bars. none of that.
just solve the two equations above, and check.

advanced:
if the equation has absolute values on both sides (i.e., |quantity1| = |quantity2| ), then you don't even have to check the solutions. they will ALWAYS work.
...but it's easier on your memory just to check them all the time - it only takes a few seconds - rather than to worry about when you have to check and when you don't.

--

so, for statement 2:
just solve
x = 3x - 2
and
x = -(3x - 2)

the solution to the former is 1. the solution to the latter is 1/2.

now, check the answers IN THE ORIGINAL EQUATION.
1 works.
1/2 does not work (the left side is positive; the right side is negative).

so, x = 1.

for once, it really is that simple!
refreshing, in a sea of problems that tend to be less simple than they appear.

[RonPurewal]

You used different approaches to check whether x=1/2 meets the requirements. First, by plugging it into the original equation 3x-2=|x| (including the absolute expression) and second, by comparing it against the region x < 0 (the region that satisfies the absolute expression) 3x-2 = -x.

So would it be correct to use either of these methods to ensure that a value of a variable (x=1/2) meets the equation requirements?

let's keep it simple.

the ONLY reliable way to CHECK AN ANSWER is to PLUG IT INTO THE ORIGINAL EQUATION.
any other method runs the risk of being compromised by little nuances that you didn't think about (not to mention the possibility that you made a mistake along the way).

if you think about it, it's pretty silly to do anything else - because, remember, the whole point here is to find solutions to the original equation.