absolute value

[jainrahul1985]

Experts please help me understand following questions using number line. I am having hard time understanding absolute value questions .

I couldn't understand how to make the cases

|x-2|+|x-3|<2

Case1: x>3

x-2 + x-3 < 2
2x < 7
x < 7/2

Case 2: 2 < x < 3

x-2 + 3 - x < 2
1 < 2

Case 3: x < 2

2 - x + 3 - x < 2
5 -2x < 2
x > 3/2


Now how to make cases for this question

|x+3|-|4-x|=|8-x|.How many questions does the equation have ?

Please explain both questions?

[/u]

[Anurag@Gurome]

Note that the given inequality has two absolute value expressions, both in terms of the variable x.

|x - 2| + |x - 3| < 2

We can see the critical points of the two absolute value expressions on a number line. The critical point of the expression |x - 2| is 2 and the critical point of the expression |x - 3| is 3. So, distinct regions on the number line with the points 2 and 3 as boundaries are: x < 2, 2 < x < 3, and x > 3. (Note that any other scenario like 3 < x < 2 is not possible at all).

Case 1: x < 2
In this case both (x - 2) and (x - 3) are negative, so the given inequality becomes:
-(x - 2) - (x - 3) < 2
-x + 2 - x + 3 < 2
5 - 2x < 2
3 < 2x
x > 3/2

Case 2: 2 < x < 3 implies (x - 2) is positive and (x - 3) is negative, so the given inequality becomes:
(x - 2) - (x - 3) < 2
1 < 2

Case 3: x > 3 implies both (x - 2) and (x - 3) are positive, so the given inequality becomes:
(x - 2) + (x - 3) < 2
2x - 5 < 2
2x < 7
x < 7/2

[vishal.pathak]

Note that the given inequality has two absolute value expressions, both in terms of the variable x.

|x - 2| + |x - 3| < 2

We can see the critical points of the two absolute value expressions on a number line. The critical point of the expression |x - 2| is 2 and the critical point of the expression |x - 3| is 3. So, distinct regions on the number line with the points 2 and 3 as boundaries are: x < 2, 2 < x < 3, and x > 3. (Note that any other scenario like 3 < x < 2 is not possible at all).

Case 1: x < 2
In this case both (x - 2) and (x - 3) are negative, so the given inequality becomes:
-(x - 2) - (x - 3) < 2
-x + 2 - x + 3 < 2
5 - 2x < 2
3 < 2x
x > 3/2

Case 2: 2 < x < 3 implies (x - 2) is positive and (x - 3) is negative, so the given inequality becomes:
(x - 2) - (x - 3) < 2
1 < 2

Case 3: x > 3 implies both (x - 2) and (x - 3) are positive, so the given inequality becomes:
(x - 2) + (x - 3) < 2
2x - 5 < 2
2x < 7
x < 7/2

Thanks Anurag.
Can you please tell the possible cases for the following inequality
|x-2| - |x-3| > |x-5|

[GMATGuruNY]

|x+3|-|4-x|=|8-x|.How many questions does the equation have ?

I received a PM asking me to discuss the equation above.

The critical points are -3, 4 and 8.
These are the values where the expressions within the absolute values are equal to 0.
To each side of these critical points, some of the expressions in the absolute values might be negative while others are positive.
If an expression will be negative with the given range, we need to flip its signs.

x<-3:
Since |x+3|<0 in this range, so we have to flip the signs in this expression.
-x-3 - (4-x) = 8-x
-7 = 8-x
x=15.
Since only values that such x<-3 are valid in this range, x=15 is not a valid solution here.

-3<x<4:
Since none of the expressions is less than 0 in this range, no signs need to be flipped.
x+3 - (4-x) = 8-x
2x-1 = 8-x
x=3.

4<x<8:
Since |4-x|<0 in this range, we have to flip the signs in this expression.
x+3 - (-4+x) = 8-x
7 = 8-x
x=1.
Since only values such that 4<x<8 are valid in this range, x=1 is not a valid solution here.

x>8:
Since |4-x|<0 and |8-x|<0 in this range, we have to flip the signs in these expressions.
x+3 - (-4+x) = -8+x
7 = -8+x
x=15.
(We can save time by recognizing that flipping the signs of one or more expressions is the equivalent of flipping the signs of the OTHER expressions.
Thus, flipping the signs of |4-x| and |8-z| will yield the same result as did flipping the signs of |x+3|.
Thus, the solution in both cases will be the same: x=15, which is valid for x>8.)

The solutions of the equation are x=3 and x=15.