Instructor help plz! - concept - abs value on both sides

[mokap25]

Hi Instructors

I am slightly fuzzy on the concept of what happens when you have multiple absolute values on both sides of an inequality equation.

I know from an earlier post from Ron Purewal that..quoting :

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.

But what if there was a constant or some other value like

Case 1 -

| QUANTITY1 | = | QUANTITY2 | + 5

Do we still do the above with 5 included as is...

Case 2
if we have
| QUANTITY1 | = | QUANTITY2 | + | QUANTITY3 |


If you can solve (or provide a link) to some questions to explain the above concepts that would be tremendously helpful.

For e.g here is a question which has 3 absolute values...if you could help solve.

Question :-

Is |a| + |b| > |a + b| ?

(1) a^2 > b^2

(2) |a| X b < 0

OA below
......


OA -E

Sorry for the long post but I really want to get my concepts here on this totally clear

Would be grateful if you could shine some light,

Thanks,
Kaps

[mithunsam]

If you have n absolute values in your equation, you would have 2^n possibilities. However, in some scenarios, some or half of them could be duplicates. So we don't have to check those possibilities.

For example, take the case |Q1| = |Q2|. There are two absolute values. Therefore, we should be having 2^2 = 4 scenarios. But, half of them will be duplicates.

Following are our possibilities

Q1 Q2 => |Q1| = |Q2|
+ + => Q1 = Q2
+ - => Q1 = -Q2
- + => -Q1 = Q2 => Q1 = -Q2 [Same as 2nd case]
- - => -Q1 = -Q2 => Q1 = Q2 [Same as 1st case]

Therefore, we can ignore 3rd and 4rth cases.

Now consider your Case 2, |Q1| = |Q2| + |Q3|. Since there are 3 absolute values, there should be 2^3 = 8 scenarios. But, half of them will be duplicates

Q1 Q2 Q3 => |Q1| = |Q2| + |Q3|
+ + + => Q1 = Q2 + Q3
+ + - => Q1 = Q2 - Q3
+ - + => Q1 = -Q2 + Q3 => Q1 = Q3 - Q2
- + + => -Q1 = Q2 + Q3 => Q1 = -Q2 - Q3
+ - - => Q1 = -Q2 - Q3 [Same as Eq 4]
- + - => -Q1 = Q2 - Q3 => Q1 = Q3 - Q2 [Same as Eq 3]
- - + => -Q1 = -Q2 + Q3 => Q1 = Q2 - Q3 [Same as Eq 2]
- - - => -Q1 = -Q2 - Q3 => Q1 = Q2 + Q3 [Same as Eq 1]

So, you have only 4 cases (first 4).

What you have to do is to multiply your current equation with -1. If you have already considered the resultant equation, then ignore your current scenario.

However, you may not get duplicates when you have a non-absolute value/variable in your equation.

Now, consider your Case 1, |Q1| = |Q2| + 5

Q1 Q2 => |Q1| = |Q2| + 5
+ + => Q1 = Q2 + 5
+ - => Q1 = -Q2 + 5 => Q1 = 5 - Q2
- + => -Q1 = Q2 + 5 => Q1 = -Q2 - 5
- - => -Q1 = -Q2 + 5 => Q1 = Q2 - 5

We got 4 different cases. This is because, the sign of 5 is static through out the cases.

[JohnHarris]

Hi,

As Ron pointed out, you can go through the different conditions but sometimes that is the long way around as mithunsam pointed out in the discussion of how the number of conditions could expand. What you would really like to do is to develop a set of tools to solve various problems. The only way to do that is to work problems, probably though reading about them or, in some cases, working through the problems. For example take your example problem:

Is |a| + |b| > |a + b| ?

(1) a^2 > b^2

(2) |a| X b < 0

We could go through the various conditions of a, b, and a+b but there is an easier way, at least IMO.

(1) Recognizing that if both a and b are of the same sign or ab=0, then
|a| + |b| = |a + b|
we let a and b be any numbers of the same sign (or ab = 0) satisfying (1). The inequality is false. (1) is not sufficient.

(2) What it really says is b is less than zero and a is not zero. Well, look at (1) and choose numbers of the same sign, i.e. both negative (since b is negative). The inequality is false. (2) is not sufficient.

(1) & (2) Look at (2) and choose two negative numbers satisfying (1). The inequality is false. (1) and (2) together is not sufficient.

[JohnHarris]

Hi,

As Ron pointed out, you can go through the different conditions but sometimes that is the long way around as mithunsam pointed out in the discussion of how the number of conditions could expand. What you would really like to do is to develop a set of tools to solve various problems. The only way to do that is to work problems, probably though reading about them or, in some cases, working through the problems. For example take your example problem:

Is |a| + |b| > |a + b| ?

(1) a^2 > b^2

(2) |a| X b < 0

We could go through the various conditions of a, b, and a+b but there is an easier way, at least IMO.

(1) Recognizing that if both a and b are of the same sign or ab=0, then
|a| + |b| = |a + b|
we let a and b be any numbers of the same sign (or ab = 0) satisfying (1). The inequality is false. (1) is not sufficient.

(2) What it really says is b is less than zero and a is not zero. Well, look at (1) and choose numbers of the same sign, i.e. both negative (since b is negative). The inequality is false. (2) is not sufficient.

(1) & (2) Look at (2) and choose two negative numbers satisfying (1). The inequality is false. (1) and (2) together is not sufficient.

[jnelson0612]

Is |a| + |b| > |a + b| ?

(1) a^2 > b^2

(2) |a| X b < 0

My way is to plug numbers:

1) Let's say a=4, b=3. This fits statement 1. Plugging these into the question in the stem, the answer is NO.
Let's change our numbers to be a=-4, b=3. Plugging into the question, the answer is YES. Insufficient.

2) b must be negative. Let's say a=4, b=-3. YES
Let's say a=-4, b=-3. NO Insufficient.

1 and 2) b must be negative, and a^2 must be greater than b^2.
a=-4, b=-3 NO
a=4, b=-3 YES Insufficient. Answer must be E.