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

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

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

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

BTW, I believe the solution should read:

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

Jeff