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:
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:
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:
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:
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
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 |
• 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 |
• 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 |
• 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 |
• 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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