Frank Solutions for Chapter 5 Linear Inequations Class 10 ICSE Mathematics
Exercise 5
1. Solve for x in the following in equations, if the replacement set is N<10:
(i) x + 5 > 11
(ii) 2x + 1 < 17
(iii) 3x – 5 ≤ 7
(iv) 8 – 3x ≥ 2
(v) 5 – 2x < 11
Answer
(i) x + 5 > 11
By transposing we get,
x > 11 – 5
⇒ x > 6
As per the condition given in the question, {x : x ∈ N;N < 10}
Therefore, solution set x = {7, 8, 9}
(ii) 2x + 1 < 17
2x + 1 < 17
By transposing we get,
2x < 17 – 1
⇒ x < 16/2
⇒ x < 8
As per the condition given in the question, {x : x ∈ N;N < 10}
Therefore, solution set x = {1, 2, 3, 4, 5, 6, 7}
(iii) 3x – 5 ≤ 7
3x – 5 ≤ 7
By transposing we get,
3x ≤ 7 + 5
⇒ x ≤ 12/3
⇒ x ≤ 4
As per the condition given in the question, {x : x ∈ N;N < 10}
Therefore, solution set x = {1, 2, 3, 4}
(iv) 8 – 3x ≥ 2
8 – 3x ≥ 2
By transposing we get,
3x ≥ 8 – 2
⇒ 3x ≥ 6
⇒ x ≥ 6/3
⇒ x ≥ 2
As per the condition given in the question, {x : x ∈ N;N < 10}
Therefore, solution set x = {1, 2}
(v) 5 – 2x < 11
5 – 2x < 11
By transposing we get,
2x > 5 – 11
⇒ 2x > -6
⇒ x > -6/2
⇒ x > -3
As per the condition given in the question, {x : x ∈ N;N < 10}
Therefore, solution set x = {1, 2, 3, 4, 5, 6, 7, 8, 9}
2. Solve for x in the following in-equations, if the replacement set is R;
(i) 3x > 12
(ii) 2x – 3 > 7
(iii) 3x + 2 ≤ 11
(iv) 14 – 3x ≥ 5
(v) 7x + 11 > 16 – 3x
(vii) 2(3x – 5) ≤ 8
(viii) x + 7 ≥ 15 + 3x
(ix) 2x – 7 ≥ 5x + 8
(x) 9 – 4x ≤ 15 – 7x
Answer
(i) 3x > 12
By cross multiplication we get,
x > 12/3
x > 4
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x > 4}
(ii) 2x – 3 > 7
2x – 3 > 7
By transposing we get,
2x > 7 + 3
⇒ 2x > 10
⇒ x > 10/2
⇒ x > 5
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x > 5}
(iii) 3x + 2 ≤ 11
3x + 2 ≤ 11
By transposing we get,
3x ≤ 11 – 2
⇒ 3x ≤ 9
⇒ x ≤ 9/3
⇒ x ≤ 3
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ 3}
(iv) 14 – 3x ≥ 5
14 – 3x ≥ 5
By transposing we get,
3x ≤ 14 – 5
⇒ 3x ≤ 9
⇒ x ≤ 9/3
⇒ x ≤ 3
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ 3}
(v) 7x + 11 > 16 – 3x
7x + 11 > 16 – 3x
By transposing we get,
7x + 3x > 16 – 11
⇒ 10x > 5
⇒ x > 5/10
⇒ x > ½
⇒ x > 0.5
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x > 0.5}
(vi) 3x + 25 > 8x – 10
3x + 25 > 8x – 10
By transposing we get,
8x – 3x < 25 + 10
⇒ 5x < 35
⇒ x < 35/5
⇒ x < 7
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x < 7}
(vii) 2(3x – 5) ≤ 8
2(3x – 5) ≤ 8
6x – 10 ≤ 8
By transposing we get,
6x ≤ 8 + 10
⇒ 6x ≤ 18
⇒ x ≤ 18/6
⇒ x ≤ 3
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ 3}
(viii) x + 7 ≥ 15 + 3x
x + 7 ≥ 15 + 3x
By transposing we get,
3x – x ≤ 7 – 15
⇒ 2x ≤ -8
⇒ x ≤ -8/2
⇒ x ≤ -4
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ -4}
(ix) 2x – 7 ≥ 5x + 8
2x – 7 ≥ 5x + 8
By transposing we get,
5x – 2x ≤ – 8 – 7
⇒ 3x ≤ – 15
⇒ x ≤ – 15/3
⇒ x ≤ – 5
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ – 5}
(x) 9 – 4x ≤ 15 – 7x
9 – 4x ≤ 15 – 7x
By transposing we get,
7x – 4x ≤ 15 – 9
⇒ 3x ≤ 6
⇒ x ≤ 6/3
⇒ x ≤ 2
As per the condition given in the question, the replacement set is R.
Therefore, solution set x = {x : x ∈ R; x ≤ 2}
3. Solve for x: 6 – 10x < 36, x ∈ {-3, -2, -1, 0, 1, 2}
Answer
From the question it is given that,
6 – 10x < 36
So, by transposing we get,
–10x < 36 – 6
⇒ –10x < 30
⇒ 10x > -30
⇒ x > – 30/10
⇒ x > – 3
As per the condition given in the question, x ∈ {-3, -2, -1, 0, 1, 2}.
Therefore, solution set x = {-2, -1, 0, 1, 2}
4. Solve for x: 3 – 2x ≥ x – 12, x ∈ N
Answer
From the question it is given that,
3 – 2x ≥ x – 12
So, by transposing we get,
2x + x ≤ 12 + 3
⇒ 3x ≤ 15
⇒ 3x ≤ 15
⇒ x ≤ 15/3
⇒ x ≤ 5
As per the condition given in the question, x ∈ N.
Therefore, solution set x = {1, 2, 3, 4, 5}
5. Solve for x : 5x – 9 ≤ 15 – 7x, x ∈ W.
Answer
From the question it is given that,
5x – 9 ≤ 15 – 7x
So, by transposing we get,
5x + 7x ≤ 15 + 9
⇒ 12x ≤ 24
⇒ x ≤ 24/12
⇒ x ≤ 2
As per the condition given in the question, x ∈ W.
Therefore, solution set x = {0, 1, 2}
6. Solve for x : 7 + 5x > x – 13, where x is a negative integer.
Answer
From the question it is given that,
7 + 5x > x – 13
So, by transposing we get,
5x – x > -13 – 7
⇒ 4x > – 20
⇒ x > -20/4
⇒ x > -5
As per the condition given in the question, x is a negative integer.
Therefore, solution set x = {-4, -3, -2, -1}
7. Solve for x: 5x – 14 < 18 – 3x, x ∈ W.
Answer
From the question it is given that,
5x – 14 < 18 – 3x
So, by transposing we get,
5x + 3x < 18 +14
⇒ 8x < 32
⇒ x < 32/8
⇒ x < 4
As per the condition given in the question, x is x ∈ W.
Therefore, solution set x = {0, 1, 2, 3}
8. Solve for x : 2x + 7 ≥ 5x – 14, where x is a positive prime number.
Answer
From the question it is given that,
2x + 7 ≥ 5x – 14
So, by transposing we get,
5x – 2x ≤ 14 + 7
⇒ 3x ≤ 21
⇒ 3x ≤ 21
⇒ x ≤ 21/3
⇒ x ≤ 7
As per the condition given in the question, x is a positive prime number.
Therefore, solution set x = {2, 3, 5, 7}
9. Solve for x : x/4 + 3 ≤ x/3 + 4, where x is a negative odd number.
Answer
From the question it is given that,
x/4 + 3 ≤ x/3 + 4
So, by transposing we get,
x/4 – x/3 ≤ 4 – 3
⇒ (3x – 4x)/12 ≤ 1
⇒ -x ≤ 12
⇒ x ≥ -12
As per the condition given in the question, x is a negative odd number.
Therefore, solution set x = {-11, -9, -7, -5, -3, -1}
10. Solve for x : (x + 3)/3 ≤ (x + 8)/ 4, where x is a positive even number.
Answer
From the question it is given that,
(x + 3)/3 ≤ (x + 8)/ 4
So, by cross multiplication we get,
4(x + 3) ≤ 3(x + 8)
⇒ 4x + 12 ≤ 3x + 24
Now, transposing we get
4x – 3x ≤ 24 – 12
⇒ x ≤ 12
As per the condition given in the question, x is a positive even number.
Therefore, solution set x = {2, 4, 6, 8, 10, 12}
11. If x + 17 ≤ 4x + 9, find the smallest value of x, when:
(i) x ∈ Z
(ii) x ∈ R
Answer
(i) x ∈ Z
From the question,
x + 17 ≤ 4x + 9
So, by transposing we get,
4x – x ≥ 17 – 9
⇒ 3x ≥ 8
⇒ x ≥ 8/3
As per the condition given in the question, x ∈ Z.
Therefore, smallest value of x = {3}
(ii) x ∈ R
From the question,
x + 17 ≤ 4x + 9
So, by transposing we get,
4x – x ≥ 17 – 9
⇒ 3x ≥ 8
⇒ x ≥ 8/3
As per the condition given in the question, x ∈ R.
Therefore, smallest value of x = {8/3}
12. If (2x + 7)/3 ≤ (5x + 1)/4, find the smallest value of x, when:
(i) x ∈ R
(ii) x ∈ Z
Answer
(i) x ∈ R
From the question,
(2x + 7)/3 ≤ (5x + 1)/4
So, by cross multiplication we get,
4(2x + 7) ≤ 3(5x + 1)
⇒ 8x + 28 ≤ 15x + 3
Now transposing we get,
15x – 8x ≥ 28 – 3
⇒ 7x ≥ 25
⇒ x ≥ 25/7
As per the condition given in the question, x ∈ R.
Therefore, smallest value of x = {25/7 }
(ii) x ∈ Z
From the question,
(2x + 7)/3 ≤ (5x + 1)/4
So, by cross multiplication we get,
4(2x + 7) ≤ 3(5x + 1)
⇒ 8x + 28 ≤ 15x + 3
Now transposing we get,
15x – 8x ≥ 28 – 3
⇒ 7x ≥ 25
⇒ x ≥ 25/7
As per the condition given in the question, x ∈ Z.
Therefore, smallest value of x = {4}
13. Solve the following linear in-equations and graph the solution set on a real number line.
(i) 2x – 11 ≤ 7 – 3x, x ∈ N.
(ii) 3(5x + 3) ≥ 2(9x – 17), x ∈ W.
(iii) 2(3x – 5) > 5(13 – 2x), x ∈ W.
(iv) 3x – 9 ≤ 4x – 7 < 2x + 5, x ∈ R.
(v) 2x – 7 < 5x + 2 ≤ 3x + 14, x ∈ R.
(vi) – 3 ≤ ½ – (2x/3) ≤ 2.2/3, x ∈ N.
(vii) 4¾ ≥ x + 5/6 > 1/3, x ∈ R
(viii) 1/3 (2x – 1) < ¼ (x + 5) < 1/6 (3x + 4), x ∈ R.
(ix) 1/3(5x – 8) ≥ ½ (4x – 7), x ∈ R.
(x) 5/4x > 1 + 1/3 (4x – 1), x ∈ R.
Answer
(i) 2x – 11 ≤ 7 – 3x
By transposing we get,
2x + 3x ≤ 7 + 11
⇒ 5x ≤ 18
⇒ x ≤ 18/5
⇒ x ≤ 3.6
As per the condition given in the question, x ∈ N.
Therefore, solution set x = {1, 2, 3}
Set can be represented in number line as,
(ii) 3(5x + 3) ≥ 2(9x – 17), x ∈ W.
From the question it is given that,
3(5x + 3) ≥ 2(9x – 17)
⇒ 15x + 9 ≥ 18x – 34
So, by transposing we get,
18x – 15x ≤ 34 + 9
⇒ 3x ≤ 43
⇒ x ≤ 43/3
As per the condition given in the question, x ∈ W.
Therefore, solution set x ≤ 43/3
Set can be represented in number line as,
(iii) 2(3x – 5) > 5(13 – 2x), x ∈ W.
From the question it is given that,
2(3x – 5) > 5(13 – 2x)
⇒ 6x – 10 > 65 – 10x
So, by transposing we get,
6x + 10x > 65 + 10
⇒ 16x > 75
⇒ x > 75/16
⇒ x > 4.6875
As per the condition given in the question, x ∈ W.
Therefore, solution set x > 4.6875
Set can be represented in number line as,
From the question,
Consider 3x – 9 ≤ 4x – 7
So, by transposing we get,
4x – 3x ≥ -9 + 7
⇒ x ≥ -2
Now, consider 4x – 7 < 2x + 5
By transposing we get,
4x – 2x < 5 + 7
⇒ 2x < 12
⇒ x < 12/2
⇒ x < 6
As per the condition given in the question, x ∈ R.
Therefore, solution set = [-2 ≤ x < 6]
Set can be represented in number line as,
(v) 2x – 7 < 5x + 2 ≤ 3x + 14, x ∈ R.
From the question,
Consider 2x – 7 < 5x + 2
By transposing we get,
5x – 2x > – 7 – 2
⇒ 3x < – 9
⇒ x < -9/3
⇒ x < -3
Now, consider 5x + 2 ≤ 3x + 14
So, by transposing we get,
5x – 3x ≤ 14 – 2
⇒ 2x ≤ 12
⇒ x ≤ 12/2
⇒ x ≤ 6
As per the condition given in the question, x ∈ R.
Therefore, solution set = [-3 < x ≤ 6]
Set can be represented in number line as,
(vi) – 3 ≤ ½ – (2x/3) ≤ 2.2/3 x ∈ N.
From the question,
Consider – 3 ≤ ½ – (2x/3)
-3 ≤ (3 – 4x)/6
⇒ –18 ≤ (3 – 4x)
So, by transposing we get,
–18 – 3 ≤ –4x
⇒ –21 ≤ –4x
⇒ x ≤ 21/4
⇒ x ≤ 5¼
Now, consider ½ – (2x/3) ≤ 2.2/3
(3 – 4x)/6 ≤ 8/3
By cross multiplication we get,
3 (3 – 4x) ≤ 48
⇒ 9 – 12x ≤ 48
By transposing we get,
–12x ≤ 48 – 9
⇒ –12x ≤ 39
⇒ 12x ≥ – 39
⇒ x ≥ – 39/12
⇒ x ≥ -3¼
As per the condition given in the question, x ∈ N.
Therefore, solution set = [-3¼ ≤ x ≤ 5¼]
Set can be represented in number line as
(vii) 4¾ ≥ x + 5/6 > 1/3, x ∈ R
From the question,
Consider, 4¾ ≥ x + 5/6
19/4 ≥ (6x + 5)/6
⇒ 114 ≥ 24x + 20
By transposing we get,
114 – 20 ≥ 24x
⇒ 94 ≥ 24x
⇒ x ≤ 94/24
⇒ x ≤ 3.92
Now, consider x + 5/6 > 1/3
(6x + 5)/6 > 1/3
⇒ 18x + 15 > 6
By transposing we get,
18x > 6 – 15
⇒ 18x > – 9
⇒ x > – 9/18
⇒ x > -½
As per the condition given in the question, x ∈ R.
Therefore, solution set = [- ½ < x ≤ 3.92]
Set can be represented in number line as
(viii) 1/3 (2x – 1) < ¼ (x + 5) < 1/6 (3x + 4), x ∈ R.
From the question it is given that,
Consider 1/3 (2x – 1) < ¼ (x + 5)
By cross multiplication we get,
4(2x – 1) < 3(x + 5)
⇒ 8x – 4 < 3x + 15
By transposing we get,
8x – 3x < 15 + 4
⇒ 5x < 19
⇒ x < 19/5
⇒ x < 3.8
Then, consider ¼ (x + 5) < 1/6 (3x + 4)
⇒ 6(x + 5) < 4(3x + 4)
⇒ 6x + 30 < 12x + 16
By transposing we get,
6x – 12x < 16 – 30
⇒ – 6x < – 14
⇒ x > 2.1/3
As per the condition given in the question, x ∈ R.
Therefore, solution set = [ 2.1/3 < x < 3.4/5]
Set can be represented in number line as
(ix) 1/3(5x – 8) ≥ ½ (4x – 7), x ∈ R.
From the question it is given that,
1/3(5x – 8) ≥ ½ (4x – 7)
By cross multiplication we get,
2(5x – 8) ≥ 3(4x – 7)
⇒ 10x – 16 ≥ 12x – 21
Transposing we get,
12x – 10x ≤ 21 – 16
⇒ 2x ≤ 5
⇒ x ≤ 5/2
⇒ x ≤ 2½
As per the condition given in the question, x ∈ R.
Therefore, solution set = {x < – 8}
Set can be represented in number line as
From the question,
Consider, (5/4)x > 1 + 1/3 (4x – 1)
⇒ (5/4)x > (3 +(4x – 1)/3)
⇒ 15x > 12 + 16x – 4
By transposing we get,
15x – 16x > 8
⇒ – x > 8
⇒ x < – 8
As per the condition given in the question, x ∈ R.
Therefore, solution set = {x < – 8}
Set can be represented in number line as,
14. If P = {x : -3 < x ≤ 7, x ∈ R} and Q = {x : – 7 ≤ x < 3, x ∈ R}, represent the following solution set on the different number lines:
(i) P ⋂ Q
(ii) Q’ ⋂ P
(iii) P – Q
Answer
As per the condition given in the question,
P = {x : -3 < x ≤ 7, x ∈ R}
So, P = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}
Then, Q = {x : – 7 ≤ x < 3, x ∈ R}
Q = {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}
(i) P ⋂ Q = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7} ⋂ {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}
= {-2, -1, 0, 1, 2}
(ii) Q’ ⋂ P
Q’ = {3, 4, 5, 6, 7}
Q’ ⋂ P = {3, 4, 5, 6, 7} ⋂ {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}
= {3, 4, 5, 6, 7}
(iii) P – Q
P – Q = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7} – {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}
= {3, 4, 5, 6, 7}
15. If P = {x : 7x – 4 > 5x + 2, x ∈ R} and Q = {x : x – 19 ≥ 1 – 3x, x ∈ R}, represent the following solution set on the different number lines:
(i) P ⋂ Q
(ii) P’ ⋂ Q
Answer
As per the condition given in the question,
P = { x : 7x – 4 > 5x + 2, x ∈ R}
7x – 4 > 5x + 2
By transposing we get,
7x – 5x > 4 + 2
⇒ 2x > 6
⇒ x > 6/2
⇒ x > 3
Therefore, P = {4, 5, 6, 7, ….}
Q = { x : x – 19 ≥ 1 – 3x, x ∈ R}
x – 19 ≥ 1 – 3x
By transposing we get,
x + 3x ≥ 1 + 19
⇒ 4x ≥ 20
⇒ x ≥ 20/4
⇒ x ≥ 5
Q = {5, 6, 7, 8, …}
Then,
(i) P ⋂ Q = {2, 3, 4, 5, ….} ⋂ {5, 6, 7, 8, …}
= {5, 6, 7, 8, …}
(ii) P’ ⋂ Q = {∅}