Chapter 22.6 - Distance as Absolute value - Solved examples - 2

In the previous section we saw solved examples which demonstrate the expression of distance as absolute value. In this section we will see a few more solved examples.

Solved example 22.10
Find the solutions to the equation: |x-1| = |x-3|
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
3. We have a point C on the number line. The number corresponding to the point C is 3
4. The difference between the numbers corresponding to A and B = (x-1)
5. Applying theorem 22.5, the distance between A and B is |(x-1)|
6. The difference between the numbers corresponding to A and C = (x-3)
7. Applying theorem 22.5, the distance between A and B is |(x-3)|
8. But these distances are given as equal.
9. That means the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.26 below:

Fig.22.26

10. If the distance from B toA is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = 1+3 = 4
• Number corresponding to the midpoint A = Half of the sum = 4/2 = 2
Thus we get x = 2
13. Check: |x-1| = |x-3| ⇒ |2-1| = |2-3| ⇒ |1| = |-1| ⇒ 1 = 1 Which is true

Solved example 22.11
Find the solutions to the equation: |x-3| = |x-4|
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 3
3. We have a point C on the number line. The number corresponding to the point C is 4
4. The difference between the numbers corresponding to A and B = (x-3)
5. Applying theorem 22.5, the distance between A and B is |(x-3)|
6. The difference between the numbers corresponding to A and C = (x-4)
7. Applying theorem 22.5, the distance between A and B is |(x-4)|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.27 below:

Fig.22.27

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = 3+4 = 7
• Number corresponding to the midpoint A = Half of the sum = 7/2 = 3.5
Thus we get x = 3.5
13. Check: |x-3| = |x-4| ⇒ |3.5-3| = |3.5-4| ⇒ |0.5| = |-0.5| ⇒ 0.5 = 0.5 Which is true

Solved example 22.12
Find the solutions to the equation: |x+2| = |x-5|
Solution:
• (x+2) can be modified as [x-(-2)]. So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -2
3. We have a point C on the number line. The number corresponding to the point C is 5
4. The difference between the numbers corresponding to A and B = [x-(-2)]
5. Applying theorem 22.5, the distance between A and B is |[x-(-2)]|
6. The difference between the numbers corresponding to A and C = (x-5)
7. Applying theorem 22.5, the distance between A and B is |(x-5)|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.27 below:

Fig.22.27

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = -2+5 = 3
• Number corresponding to the midpoint A = Half of the sum = 3/2 = 1.5
Thus we get x = 1.5
13. Check: |x+2| = |x-5| ⇒ |1.5+2| = |1.5-5| ⇒ |3.5| = |-3.5| ⇒ 3.5 = 3.5 Which is true

Solved example 22.13
Find the solutions to the equation: |x| = |x+1|
Solution:
• |x| is the distance of the 'point whose number is x' from zero
• (x+1) can be modified as [x-(-1)]. 
So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
3. The distance of A from zero is |x|
4. The difference between the numbers corresponding to A and B =  [x-(-1)]
5. Applying theorem 22.5, the distance between A and B is | [x-(-1)]|
6. But these distances are given as equal.
7. That means, the distance from B to A is same as the distance from A to zero. Now look at the number line in fig.22.28 below:

Fig.22.28

8. If the distance from B to A is same as the distance from A to zero, there is only one possibility:
• A is the midpoint between B and zero
9. So we are in the following situation:
• We have a point B, whose number is known
• We have the point zero, whose number is '0'
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and zero
10. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and zero = -1+0 = -1
• Number corresponding to the midpoint A = Half of the sum = -1/2 = -0.5
Thus we get x = -0.5
11. Check: |x| = |x+1| ⇒ |-0.5| = |-0.5+1| ⇒ |-0.5| = |0.5| ⇒ 0.5 = 0.5 Which is true

Alternate method:
• x can be modified as (x-0)
• (x+1) can be modified as [x-(-1)]. 
So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 0
3. We have a point C on the number line. The number corresponding to the point C is -1
4. The difference between the numbers corresponding to A and B = (x-0)
5. Applying theorem 22.5, the distance between A and B is |(x-0)|
6. The difference between the numbers corresponding to A and C =  [x-(-1)]
7. Applying theorem 22.5, the distance between A and B is | [x-(-1)]|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.29 below:

Fig.22.29

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = -1+0 = -1
• Number corresponding to the midpoint A = Half of the sum = -1/2 = -0.5
Thus we get x = -0.5
13. Check: |x| = |x+1| ⇒ |-0.5| = |-0.5+1| ⇒ |-0.5| = |0.5| ⇒ 0.5 = 0.5 Which is true

Solved example 22.14
Prove that if 1 < x < 4 and 1 < y < 4, then |x-y| < 3
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
3. We have a point C on the number line. The number corresponding to the point C is 4
4. We have a point D on the number line. The number corresponding to the point D is y
5. Given that 1 < x < 4. So the point A can be any where in between C and D
x can never take a value of exact 1 or exact 4. So hollow circles are shown at B and C in the graph below:

Fig.22.30

6. Given that 1 < y < 4. So the point D can be any where in between B and C
y can never take a value of exact 1 or exact 4.
7. We want the difference between x and y. The 'maximum difference' will be obtained in the following situation:
• When one of them (x or y) is at it's maximum possible value AND
• The other is at it's minimum possible value
8. When one of them is at it's maximum possible value, it will be 'just to the left of 4'
Then the other is at the minimum possible value, which is 'just to the right of 1'
9. Concentrate on the two positions:
• One point is 'just to the left of 4'
• The other point is 'just to the right of 1'
10. In such a situation, the distance between them (calculated as |x-y|) will be less than 3
■ We will get 3 only if one is at 'exact 4' and the other is at 'exact 1'
11. Thus we find that the maximum possible distance is less than 3
We can write: |x-y| < 3

Solved example 22.15
Prove that if x < 3 and y > 7, then |x-y| > 4
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 3
3. We have a point C on the number line. The number corresponding to the point C is 7
4. We have a point D on the number line. The number corresponding to the point D is y
5. Given that x < 3. So the point A can be any where on the red line

x can never take a value of exact 3. So hollow circle is shown at B in the graph below:

Fig.22.31

6. Given that y > 7. So the point D can be any where on the blue line
y can never take a value of exact 7.
7. We want the difference between x and y. The 'minimum difference' will be obtained in the following situation:
• When x is just to the left of B AND
• y is just to the right of C
• If any one of them is at a different position other than the above two, the difference will not be minimum
8. Concentrate on the two positions:
• x is 'just to the left of B'
• y is 'just to the right of C'
10. In such a situation, the distance between them (calculated as |x-y|) will be greater than 4
■ We will get 4 only if x is at 'exact 3' and y is at 'exact 7'
11. Thus we find that the minimum possible distance is greater than 4
We can write: |x-y| > 4


Solved example 22.16
What are the numbers x, for which |x-2| + |x-8| = 6
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 2

3. We have a point C on the number line. The number corresponding to the point C is 8
4. The difference between the numbers corresponding to A and B = (x-2)
5. Applying theorem 22.5, the distance between A and B is |x-2|
6. The difference between the numbers corresponding to A and C =  (x-8)
7. Applying theorem 22.5, the distance between A and B is |x-8|
8. The sum of the two distances is given as 6. Consider the number line shown in fig.21.32 below:

Fig.22.32

9. We can see that the actual distance between B and C is (8-2) = 6 units
10. So A can lie anywhere on the red line. It can also take the exact values 2 and 8
11. So we can write 2 ≤ x ≤ 8
12. Check: • Let x = 2
Then |x-2| + |x-8| = 6 ⇒|2-2| + |2-8| = 6 ⇒ |0| + |-6| = 6 ⇒ 0 + 6 = 6 ⇒6 = 6 which is true 
• Let x = 4
Then |x-2| + |x-8| = 6 ⇒|4-2| + |4-8| = 6 ⇒ |2| + |-4| = 6 ⇒ 2 + 4 = 6 ⇒6 = 6 which is true
• Let x = 8
Then |x-2| + |x-8| = 6 ⇒|8-2| + |8-8| = 6 ⇒ |6| + |0| = 6 ⇒ 6 + 0 = 6 ⇒6 = 6 which is true

Solved example 22.17
What are the numbers x, for which |x-2| + |x-8| = 10
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 2

3. We have a point C on the number line. The number corresponding to the point C is 8
4. The difference between the numbers corresponding to A and B = (x-2)
5. Applying theorem 22.5, the distance between A and B is |x-2|
6. The difference between the numbers corresponding to A and C =  (x-8)
7. Applying theorem 22.5, the distance between A and B is |x-8|
8. The sum of the two distances is given as 10. Consider the number line shown in fig.21.33 below:

9. The two possible positions of A are shown in blue colour. The corresponding numbers are 0 and 10
10. Check: • Let x = 0
Then |x-2| + |x-8| = 10 ⇒|0-2| + |0-8| = 10 ⇒ |-2| + |-8| = 10 ⇒ 2 + 8 = 10 ⇒10 = 10 which is true   
• Let x = 10
Then |x-2| + |x-8| = 10 ⇒|10-2| + |10-8| = 10 ⇒ |8| + |2| = 10 ⇒ 8 + 2 = 10 ⇒10 = 10 which is true
Another method:
We will consider all the possible values that x can take.
Case 1: x < 2
1. Then (x-2) will be a negative quantity. This is same as (2-x) is a positive quantity
2. Whether we take (x-2) or (2-x), both are 'difference'. And when we take the absolute value, we are dropping the '-' sign.
3. So, when x < 2, |x-2| = (2-x)
4. When x is less than 2, it will be less than 8 also. So |x-8| = (8-x)
5. Substituting these values in the given equation we get:
(2-x) + (8-x) = 10 ⇒ 10 -2x = 10 ⇒ 2x = 0 ⇒ x = 0
Case 2: x = 2
1. Then (x-2) = 0 and |0| = 0
2. |x-8| = |(2-8)| = |-6| = 6
3. Substituting these values in the given equation we get:
0 + 6 = 10. This is not true. So we cannot consider case 2
Case 3: x > 2 AND x < 8
1. That means x is greater than 2, but at the same time, x is less than 8
2. Another way of writing this is: 2 <  x < 8
3. Graphically, this will be a line from 2 to 8, with hollow circles at 2 and 8
4. Let us see whether any of such values of x satisfies the given equation:
5. When x > 2, (x-2) is positive. So |x-2| = (x-2)
6. When x is less than 8, (x-8) is negative. So |x-8| = (8-x)
7. Substituting these values in the given equation we get:
(x-2) + (8-x) = 10 ⇒ 6 = 10. This is not true. So we cannot consider case 3
Case 4: x = 8
1. Then (x-2) = 6 and |6| = 6
2. |x-8| = 0 and |0| = 0
3. Substituting these values in the given equation we get:
6 + 0 = 10. This is not true. So we cannot consider case 4
Case 5: x > 8
1. Then x will be greater than 2 also. So |x-2| = (x-2)    
2. (x-8) will be positive. So |x-8| = (x-8)
3. Substituting these values in the given equation we get:    
4. (x-2) + (x-8) = 10 ⇒ 2x-10 = 10 ⇒ 2x = 20 ⇒ x = 10

■ So we considered 5 cases. Those 5 cases cover all possible numbers on the number line. But only the cases 1 and 5 will give us a valid result.
• The valid results are: x = 0 and x = 10

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Chapter 22.5 - Distance as Absolute value - Solved examples

In the previous section we saw the 'distance between any two points' expressed as absolute value. We also saw some practical cases. In this section we will see a few more cases.

The equations that we saw were:
• |x-1| = 3 
• |x+1| = 3 
Now we will see some inequalities:
Solved example 22.6
Find the solutions to the inequality |x-1| ≤ 3 
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
• The difference between the numbers = (x-1)
3. Applying theorem 22.5, the distance between A and B is |(x-1)|
4. But |(x-1)| is less than or equal to 3. So the distance between A and B should never exceed 3 units
5. That means A should always be within a distance of 3 units from B. Now look at the number line in fig.22.22 below:

Fig.22.22

6. There are two possibilities:
• A can be within a distance of 3 units towards the left from B
• A can be within a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (1 - 3) = -2
• So A is not allowed to take any position to the left of -2. If it does, it's distance from B will exceed 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (1 + 3) = 4
• So A is not allowed to take any position to the right of 4. If it does, it's distance from B will exceed 3 units
11. All the points that lie in between -2 and 4 will satisfy the given inequality. If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.22 above.
12. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points -2 and 4 also qualify to be part of the graph
13. If the inequality is |x-1| < 3, then only those points whose distances are less than 3 from B should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.22 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -1.5 is a number in the red line
Put x = -1.5 in the inequality |x-1| ≤ 3
We get: |-1.5-1| ≤ 3 ⇒ |-2.5| ≤ 3 ⇒ 2.5 ≤ 3 Which is true

Solved example 22.7
Find the solutions to the inequality |x-1| ≥ 3 
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
• The difference between the numbers = (x-1)
3. Applying theorem 22.5, the distance between A and B is |(x-1)|
4. But |(x-1)| is greater than or equal to 3. So the distance between A and B should never be less than 3 units
5. That means A should always be away from B by a distance of 3 units or more. In other words, A should never come closer to B by a distance less than 3 units. Now look at the number line in fig.22.23 below:

Fig.22.23

6. There are two possibilities:
• A can be away by a distance of 3 or more units towards the left from B
• A can be away by a distance of 3 or more units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (1 - 3) = -2
• So A is not allowed to take any position to the right of -2. If it does, it's distance from B will become less than 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (1 + 3) = 4
• So A is not allowed to take any position to the left of 4. If it does, it's distance from B will become less than 3 units
11. ♦ All the numbers that lie to the left of -2 will satisfy the given inequality.
      ♦ All the numbers that lie to the right of 4 will satisfy the given inequality.
If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.22 above.
• But the red line is broken. This is because, the points which lie in between -2 and 4 does not satisfy the given inequality
• Arrows are given at the left and right ends of the line. This is to show that the graph extends upto infinity on both sides.
    ♦ All numbers on the left of -2 upto infinity satisfies the inequality
    ♦ All numbers on the right of 4 upto infinity satisfies the inequality
12. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points -2 and 4 also qualify to be part of the graph
13. If the inequality is |x-1| > 3, then only those points whose distances are greater than 3 should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.23 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -9.3 is a number in the red line
Put x = -9.3 in the inequality |x-1| ≥ 3

We get: |-9.3-1| ≥ 3 ⇒ |-10.3| ≥ 3 ⇒ 10.3 ≥ 3 Which is true

Solved example 22.8
Find the solutions to the inequality |x+1| ≤ 3
Solution:
• (x+1) can be written as [x-(-1)]
• So  |x+1| ≤ 3 ⇒ |[x-(-1)]| ≤ 3
• Based on this modification we can write:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
• The difference between the numbers = [x-(-1)]
3. Applying theorem 22.5, the distance between A and B is |[x-(-1)]|
4. But |[x-(-1)]| is less than or equal to 3. So the distance between A and B should never exceed 3 units
5. That means A should always be within a distance of 3 units from B. Now look at the number line in fig.22.24 below:

Fig.22.24

6. There are two possibilities:
• A can be within a distance of 3 units towards the left from B
• A can be within a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (-1 - 3) = -4
• So A is not allowed to take any position to the left of -4. If it does, it's distance from B will exceed 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (-1 + 3) = 2
• So A is not allowed to take any position to the right of 2. If it does, it's distance from B will exceed 3 units
11. All the points that lie in between -4 and 2 will satisfy the given inequality. If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.24 above.
12. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points -4 and 2 also qualify to be part of the graph
13. If the inequality is |x+1| < 3, then only those points whose distances are less than 3 from B should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.24 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -3.75 is a number in the red line
Put x = -3.75 in the inequality |x+1| ≤ 3

We get: |-3.75+1| ≤ 3 ⇒ |-2.75| ≤ 3 ⇒ 2.75 ≤ 3 Which is true

Solved example 22.9
Find the solutions to the inequality |x+1| ≥ 3
Solution:
• (x+1) can be written as [x-(-1)]
• So  |x+1| ≥ 3 ⇒ |[x-(-1)]| ≥ 3
• Based on this modification we can write:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
• The difference between the numbers = [x-(-1)]
3. Applying theorem 22.5, the distance between A and B is |[x-(-1)]|
4. But |(x-1)| is greater than or equal to 3. So the distance between A and B should never be less than 3 units
5. That means A should always be away from B by a distance of 3 units or more. In other words, A should never come closer to B by a distance less than 3 units. Now look at the number line in fig.22.25 below:

Fig.22.25

6. There are two possibilities:
• A can be away by a distance of 3 or more units towards the left from B
• A can be away by a distance of 3 or more units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (-1 - 3) = -4
• So A is not allowed to take any position to the right of -4. If it does, it's distance from B will become less than 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (-1 + 3) = 2
• So A is not allowed to take any position to the left of 2. If it does, it's distance from B will become less than 3 units
11. ♦ All the numbers that lie to the left of -4 will satisfy the given inequality.
      ♦ All the numbers that lie to the right of 2 will satisfy the given inequality.
If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.25 above.
• But the red line is broken. This is because, the points which lie in between -4 and 2 does not satisfy the given inequality
• Arrows are given at the left and right ends of the line. This is to show that the graph extends upto infinity on both sides.
    ♦ All numbers on the left of -4 upto infinity satisfies the inequality
    ♦ All numbers on the right of 2 upto infinity satisfies the inequality
12. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points -4 and 2 also qualify to be part of the graph
13. If the inequality is |x+1| > 3, then only those points whose distances are greater than 3 should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.23 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -20.5 is a number in the red line
Put x = -20.5 in the inequality |x+1| ≥ 3
We get: |-20.5+1| ≥ 3 ⇒ |-19.5| ≥ 3 ⇒ 19.5 ≥ 3 Which is true

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In the next section we will see more solved examples.

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Chapter 22.4 - Distance as Absolute value

In the previous section we saw that the distance between any number and zero is the absolute value of that number. In this section we will see the 'distance between any two points' expressed as absolute value.

We have seen the method to calculate the distance between any two points on the number line:
• The smaller number is subtracted from the larger number. 
We saw it in theorem 22.1. Let us analyse it again:
Calculation of the distance between a few pairs of points is shown below. Calculations are based on theorem 22.1.

Example 1
Number corresponding to Point A = 5
Number corresponding to Point B = 3
Smaller of the two is 3
So distance between A and B = 5 – 3 = 2 units
Example 2
Number corresponding to Point A = 4
Number corresponding to Point B = -2
Smaller of the two is -2
So distance between A and B = 4 – (-2) = 4 + 2 = 6 units

Example 3
Number corresponding to Point A = -8
Number corresponding to Point B = -5
Smaller of the two is -8
So distance between A and B = -5 – (-8) = -5 + 8 = 3 units

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■ Note that, when we subtract the smaller from the larger, we are calculating the 'difference' between the two numbers. 

■ If we subtract the larger number from the smaller, then also, we will get the 'difference'. In that case, we will get a negative answer. 

Let us check for the above examples:
• In example 1, we got the result as 2 units.
If we subtract the larger from the smaller, the result will be 3 – 5 = -2 units
• In example 2, we got the result as 6 units.
If we subtract the larger from the smaller, the result will be -2 – 4 = -6 units
• In example 3, we got the result as 3 units.
If we subtract the larger from the smaller, the result will be -8 – (-5) = -8 + 5 = -3 units

So we find that the signs have reversed. But for specifying distance, we need only the magnitude. Sign does not have a role to play. For example, the petrol consumed by a car for travelling 10 km will be the same whether it travels towards the east, or towards the west, if the road conditions are the same.

That means, we do not have to strictly follow the rule:
Subtract smaller number from the larger number

All we need is the difference. If we take the absolute value of that difference, we will get the required distance without any sign. Let us see the above 3 examples again:

• In example 1, we may get the difference as 2 or -2, whatever be the difference, we must take it's absolute value:
|2| = 2 AND |-2| = 2
• In example 2, we may get the difference as 6 or -6, whatever be the difference, we must take it's absolute value:
|6| = 6 AND |-6| = 6
• In example 3, we may get the difference as 3 or -3, whatever be the difference, we must take it's absolute value:
|3| = 3 AND |-3| = 3
■ So, if we decide to always take the absolute value of the difference, we will get the required distance, regardless of whether we subtract the smaller from the larger or vice versa.


We can write this in the form of a theorem:
Theorem 22.5
• We have a point A on the number line. We know the number corresponding to the point A
• We have a point B on the number line. We know the number corresponding to the point B
• Calculate the difference between the numbers
• Take the absolute value of the difference
• This absolute value is the 'distance between the points A and B'

Example:
• We have a point A on the number line. The number corresponding to the point A is 2
• We have a point B on the number line. The number corresponding to the point B is 5
• Calculate the difference between the numbers: (2 - 5) = -3
• Take the absolute value of the difference: |-3| = 3
• This absolute value '3' is the distance between the points A and B

Now we will see some solved examples:
Solved example 22.4
• Consider the equation |x-1| = 3
• Find the value of x that will satisfy the above equation
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
3. The difference between the numbers = (x-1)
3. Applying theorem 22.5, the distance between A and B is |(x-1)|
4. But |(x-1)| is given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.20 below:

Fig.22.20

6. There are two possibilities:
• A can be at a distance of 3 units towards the left from B
• A can be at a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the distance between A and B.
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (1 - 3) = -2
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the distance between A and B.
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (1 + 3) = 4
11. • From (8), we have A = -2
• From (10), we have A = 4
• Both values will satisfy the equation |x-1| = 3 
• They are marked by red 'filled circles' in the number line in fig.22.20
12. Check: Put x = -2 in the equation |x-1| = 3
We get: |-2-1| = 3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 4 in the equation |x-1| = 3
We get: |4-1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x-1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
    ♦ Finding the solutions to the equation
    ♦ Solving the equation

Solved example 22.5
• Consider the equation |x+1| = 3
• Find the value of x that will satisfy the above equation
Solution:
• (x+1) can be written as [x-(-1)]
• So  |x+1| = 3 ⇒ |[x-(-1)]| = 3
• Based on this modification we can write:
1. We have a point A on the number line. The number corresponding to the point A is x

2. We have a point B on the number line. The number corresponding to the point B is -1
3. The difference between the numbers = [x-(-1)]
3. Applying theorem 22.5, the distance between A and B is |[x-(-1)]|
4. But |[x-(-1)]| is given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.21 below:

Fig.22.21

6. There are two possibilities:
• A can be at a distance of 3 units towards the left from B
• A can be at a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the distance between A and B.
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (-1 - 3) = -4
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the distance between A and B.
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (-1 + 3) = 2
11. • From (8), we have A = -4
• From (10), we have A = 2
• Both values will satisfy the equation |[x-(-1)]| = 3 
• They are marked by red 'filled circles' in the number line in fig.22.21
12. Check: Put x = -4 in the equation |x+1| = 3
We get: |-4+1| = -3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 2 in the equation |x+1| = 3
We get: |2+1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x+1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
    ♦ Finding the solutions to the equation
    ♦ Solving the equation

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In the above two solved examples, we considered 'equations'. In the next section we will consider 'inequalities'.

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