Worksheet on Solution of a Linear Inequation in One Variable | Solving Linear Inequalities in One Variable Worksheet PDF

Given that if x ∈ N, then
(i) 10x + 6 ≤ 4x + 36
From the given information, x is the variable.
Firstly, move the x terms on one side.
10x – 4x + 6 ≤ 36
Now, subtract 6 from both sides of the above equation.
10x – 4x + 6 – 6 ≤ 36 – 6
10x – 4x ≤ 30
6x ≤ 30
Now, divide the above equation with 6 into both sides.
6x/6 ≤ 30/6
x ≤ 5.
As x ∈ N, the solution set of the linear inequations becomes {1, 2, 3, 4, 5}

Therefore, the answer is {1, 2, 3, 4, 5}.
(ii) 6x – 4 < 38 – 8x
From the given information, x is the variable.
Firstly, move the x terms on one side.
6x + 8x – 4 < 38
Now, add 4 on both sides of the above equation.
6x + 8x – 4 + 4 < 38 + 4
6x + 8x < 42
14x < 42
Now, divide the above equation by 14 into both sides.
14x/14 < 42/14
x < 3.
As x ∈ N, the solution set of the linear inequations becomes {1, 2}

Therefore, the answer is {1, 2}.

(i) Given inequation is 8x + 6 < 6x – 2.
We need to find out x = -4 a solution of inequation 8x + 6 < 6x – 2 or not.
Now, substitute x = -4 in the inequation 8x + 6 < 6x – 2.
8x + 6 < 6x – 2 = 8(-4) + 6 < 6(-4) – 2
-32 + 6 < -24 – 2
-26 < -26 which is not correct.
x = -4 not a solution of inequation 8x + 6 < 6x – 2

As -26 < -26 is false, x = -4 not a solution of inequation 8x + 6 < 6x – 2
(ii) Given inequation is 4x + 2 ≥ 2x – 6.
We need to find out x = 2 a solution of inequation 4x + 2 ≥ 2x – 6 or not.
Now, substitute x = 2 in the inequation 4x + 2 ≥ 2x – 6.
4 (2) + 2 ≥ 2 (2) – 6
8 + 2 ≥ 4 – 6
10 ≥ -2 which is correct.
x = 2 is a solution of inequation 4x + 2 ≥ 2x – 6.

As 10 ≥ -2 is true, x = 2 is a solution of inequation 4x + 2 ≥ 2x – 6.

Given inequation is 9 – 6x ≥ 3x – 36 when x ∈ N.
From the given information, x is the variable.
Firstly, move the x terms on one side.
9 – 6x -3x ≥ – 36
Now, subtract 9 from both sides of the above equation.
9 – 6x -3x -9 ≥ – 36 – 9
– 9x ≥ – 45
Now, divide the above equation by -9 into both sides.
– 9x/-9 ≥ – 45/-9
The inequality reverses on multiplying both sides by -1.
x ≤ 5
As x ∈ N, the solution set of the linear inequations becomes {1, 2, 3, 4, 5}

Therefore, the answer is {1, 2, 3, 4, 5}.

(i) Given inequation 2x – 4 > 6 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
2x – 4 > 6
Now, add 4 on both sides of the above equation.
2x – 4 +4 > 6 + 4
2x > 10
Now, divide the above equation by 2 into both sides.
2x/2 > 10/2
x > 5
Therefore, the answer is x > 5.

(ii) Given inequation 6x < 30 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
6x < 30
Now, divide the above equation by 6 into both sides.
6x/6 < 30/6
x < 5
Therefore, the answer is x < 5.

(iii) Given inequation -12x ≥ -48 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
-12x ≥ -48
Now, divide the above equation by -12 into both sides. The inequality reverses on multiplying both sides by -1.
-12x/-12 ≥ -48/-12
x ≤ 4
Therefore, the answer is x ≤ 4.

(iv) Given inequation 16x – 12 ≥ 36 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
16x – 12 ≥ 36
Now, add 12 on both sides of the above equation.
16x – 12 + 12 ≥ 36 + 12
16x ≥ 48
Now, divide the above equation by 16 into both sides.
16x/16 ≥ 48/16
x ≥ 3
Therefore, the answer is x ≥ 3.

(v) Given inequation 10 – 4x < 230 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
10 – 4x < 230.
Now, subtract 10 from both sides of the above equation.
10 -10 – 4x < 230 – 10.
– 4x < 220.
Now, divide the above equation by -4 into both sides. The inequality reverses on multiplying both sides by -1.
– 4x/-4 < 220/-4.
x > -55
Therefore, the answer is x > -55.

Given inequation is 100 – 16x ≤ 64.
(i) From the given information, x is a real number.
Firstly, move the x terms on one side.
100 – 16x ≤ 64
Now, subtract 100 from both sides of the above equation.
100 – 100 – 16x ≤ 64 -100
– 16x ≤ -36
Now, divide the above equation by -16 into both sides. The inequality reverses on multiplying both sides by -1.
– 16x/-16 ≤ -36/-16
x ≥ \(\frac { 9 }{ 4 } \)

Therefore, the smallest value of x, when x is a real number is \(\frac { 9 }{ 4 } \).
(ii) The smallest value of x, when x is an integer is 3.

Given that x is a positive integer satisfying 90 – 12(6x + 3) < 30.
From the given information, x is a variable.
Firstly, move the x terms on one side. Simplify the given inequation.
90 – 12(6x + 3) < 30
90 – 72x – 36 < 30
54 – 72x < 30
Now, subtract 54 from both sides of the above equation.
54 – 54 – 72x < 30 – 54
-72x < -24
Now, divide the above equation by -72 into both sides. The inequality reverses on multiplying both sides by -1.
-72x/-72 < -24/-72
x > \(\frac { 1 }{ 3 } \)

Therefore, the solution set is {1, 2, 3, …..}

(i) Given inequation -2x + 14 > 8x – 6 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
-2x + 14 > 8x – 6
14 > 8x + 2x – 6
14 > 10x – 6
Now, add 6 on both sides of the above equation.
14 + 6 > 10x – 6 + 6
20 > 10x
Now, divide the above equation by 10 into both sides.
20/10 > 10x/10
x < 2
Therefore, the answer is x < 2.

(ii) Given inequation 14x – 10x ≥ 6 + 2x in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
14x – 10x ≥ 6 + 2x
4x ≥ 6 + 2x
4x – 2x ≥ 6
2x ≥ 6
Now, divide the above equation by 2 into both sides.
2x/2 ≥ 6/2
x ≥ 3
Therefore, the answer is x ≥ 3.

(iii) Given inequation 4(2x + 1) ≤ x + 10 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
4(2x + 1) ≤ x + 10
8x + 4 ≤ x + 10
Move x to left side.
8x -x + 4 ≤ 10
7x + 4 ≤ 10
Now, subtract 4 from both sides of the above equation.
7x + 4 – 4 ≤ 10 – 4
7x ≤ 6
Now, divide the above equation by 7 into both sides.
7x/7 ≤ 6/7
x ≤ \(\frac { 6 }{ 7 } \)
Therefore, the answer is x ≤ \(\frac { 6 }{ 7 } \).

(iv) Given inequation 10(6x – 4) < 6(8x – 6) in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
10(6x – 4) < 6(8x – 6)
60x – 40 < 48x – 36
Move 48x to left side.
60x – 48x – 40 < -36
12x – 40 < -36
Now, add 40 on both sides of the above equation.
12x + 40 – 40 < -36 + 40
12x < 4
Now, divide the above equation by 12 into both sides.
12x/12 < 4/12
x < \(\frac { 1 }{ 3 } \)
Therefore, the answer is x < \(\frac { 1 }{ 3 } \).

(v) Given inequation 6 + \(\frac { x }{ 2 } \) > \(\frac { 2x }{ 5 } \) + 14 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
\(\frac { 12 + x }{ 2 } \) > \(\frac { 2x + 70 }{ 5 } \)
Move 2 to the right side and multiply it with \(\frac { 2x + 70 }{ 5 } \) and move 5 to the left side and multiply it with \(\frac { 12 + x }{ 2 } \).
5 (12 + x) > 2 (2x + 70)
Solve the above equation.
60 + 5x > 4x + 140
Move 4x to the left side.
60 + 5x – 4x > 140
60 + x > 140
Now, subtract 60 from both sides of the above equation.
60 – 60 + x > 140 – 60
x > 80
Therefore, the answer is x > 80.

(vi) Given inequation \(\frac { 2x – 1 }{ 7 } \) ≥ \(\frac { 2x + 3 }{ 3 } \) in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
\(\frac { 2x – 1 }{ 7 } \) ≥ \(\frac { 2x + 3 }{ 3 } \)
Move 7 to the right side and multiply it with \(\frac { 2x + 3 }{ 3 } \) and move 3 to the left side and multiply it with \(\frac { 2x – 1 }{ 7 } \).
3 (2x – 1) ≥ 7 (2x + 3)
Solve the above equation.
6x – 3 ≥ 14x + 21
Move 6x to the right side.
– 3 ≥ 14x – 6x + 21
– 3 ≥ 8x + 21
Now, subtract 21 from both sides of the above equation.
– 3 -21 ≥ 8x + 21 -21
– 24 ≥ 8x
Now, divide the above equation by 8 into both sides.
– 24/8 ≥ 8x/8
– 3 ≥ x
Therefore, the answer is x ≤ -3.

If x and y be positive integers satisfying x + y ≤ 4.
As mentioned x and y must be positive integers, then the x value must be 1 or 2 or 3.
If x = 1, then 1 + y ≤ 4.
y ≤ 4 – 1
y = 3.
If x = 2, then 2 + y ≤ 4.
y ≤ 4 – 2
y = 2.
If x = 3, then 3 + y ≤ 4.
y ≤ 4 – 3
y = 3.

Therefore, the possible values of x and y are (1, 3), (2, 2), and (3, 1)

Given inequation is 4(x – 1) ≤ 18 – 2x and x ∈ W.
Solve the inequation 4(x – 1) ≤ 18 – 2x
4x – 4 ≤ 18 – 2x
Move -2x to the left side.
4x + 2x – 4 ≤ 18
6x – 4 ≤ 18
Now, add 4 on both sides of the above equation.
6x – 4 + 4 ≤ 18 + 4
6x ≤ 22
Now, divide 6 into both sides of the above equation.
6x/6 ≤ 22/6
x ≤ 3.6666

Therefore, the largest value of x for which 4(x – 1) ≤ 18 – 2x and x ∈ W is 3.

(i) Given inequation is 6 + 10x > 6x – 6, where x is a negative integer
Solve the inequation 6 + 10x > 6x – 6
Move 6x to the left side.
6 + 10x -6x > – 6
6 + 4x > – 6
Now, subtract 6 from both sides of the above equation.
6 + 4x – 6 > – 6 – 6
4x > – 12
Now, divide 4 into both sides of the above equation.
4x/4 > – 12/4
x > -3
Therefore, the x values become -2 and -1.

(ii) Given inequation is 10x + 8 < 6x + 40, where x ∈ N.
Solve the inequation 10x + 8 < 6x + 408.
Move 6x to the left side.
8 + 10x -6x < 40
8 + 4x < 40
Now, subtract 8 from both sides of the above equation.
8 + 4x – 8 < 40 – 8
4x < 32
Now, divide 4 into both sides of the above equation.
4x/4 < 32/4
x < 8
Therefore, the solution set of 10x + 8 < 6x + 40, where x ∈ N is {1,2,3,4,5,6,7}

(iii) Given inequation is x + 5 ≤ \(\frac { 2x }{ 3 } \) + 6, where x is a positive odd integer.
Solve the inequation x + 5 ≤ \(\frac { 2x }{ 3 } \) + 6.
Move \(\frac { 2x }{ 3 } \) to the left side.
x – \(\frac { 2x }{ 3 } \) + 5 ≤ 6
\(\frac { x }{ 3 } \) + 5 ≤ 6
Now, subtract 5 from both sides of the above equation.
\(\frac { x }{ 3 } \) + 5 – 5 ≤ 6 – 5
\(\frac { x }{ 3 } \) ≤ 1
Now, multiply 3 on both sides of the above equation.
3 × \(\frac { x }{ 3 } \) ≤ 1 × 3
x ≤ 3
As it is mentioned x is a positive odd integer, the answer becomes {1, 3}
Therefore, the answer is {1, 3}.

(iv) Given inequation 4x + 6 ≥ 2x + 10, where x is a natural number less than 8.
Solve the inequation 4x + 6 ≥ 2x + 10.
Move 2x to the left side.
4x – 2x + 6 ≥ 10
2x + 6 ≥ 10
Now, subtract 6 from both sides of the above equation.
2x + 6 – 6 ≥ 10 – 6
2x ≥ 4
Now, divide 2 into both sides of the above equation.
2x/2 ≥ 4/2
x ≥ 2
As it is mentioned x is a natural number less than 8, the answer becomes {2, 3, 4, 5, 6, 7}

(v) Given inequation is \(\frac { 2x + 6 }{ 3 } \) ≤ \(\frac { 2x + 16 }{ 4 } \), where x is positive even integer.
Solve the inequation \(\frac { 2x + 6 }{ 3 } \) ≤ \(\frac { 2x + 16 }{ 4 } \).
Move 3 to the right side and multiply it with \(\frac { 2x + 16 }{ 4 } \) and move 4 to the left side and multiply it with \(\frac { 2x + 6 }{ 3 } \).
4 (2x + 6) ≤ 3 (2x + 16)
8x + 24 ≤ 6x + 48
Move 6x to the left side.
8x – 6x + 24 ≤ 48
2x + 24 ≤ 48
Now, subtract 24 from both sides of the above equation.
2x + 24 – 24 ≤ 48 – 24
2x ≤ 24
Now, divide 2 into both sides of the above equation.
2x/2 ≤ 24/2
x ≤ 12
As it is mentioned x is positive even integer, the answer becomes {2, 4, 6, 8, 10, 12}

(vi) Given inequation is \(\frac { 6x }{ 5 } \) – \(\frac { 4 (x – 2) }{ 3 } \) > 2, where x ∈ {2, 4, 6, 8, 10}
Solve the inequation\(\frac { 6x }{ 5 } \) – \(\frac { 4 (x – 2) }{ 3 } \) > 2.
\(\frac { 6x }{ 5 } \) – \(\frac { 4x – 8) }{ 3 } \) > 2
\(\frac { 18x – 20x + 40}{ 15 } \) > 2
\(\frac { – 2x + 40}{ 15 } \) > 2
Multiply 15 on both sides of the above equation.
– 2x + 40 > 30
Now, subtract – 40 from both sides of the above equation.
– 2x + 40 – 40 > 30 -40
– 2x > -10
Now, divide -2 into both sides of the above equation. The inequality reverses on multiplying both sides by -1.
2x/2 < 10/2
x < 5
As it is mentioned x ∈ {2, 4, 6, 8, 10}, the answer becomes {2, 4}

Given inequation -24x ≥ -96 in R.
From the given information, x is the variable.
Firstly, move the x terms on one side.
-24x ≥ -96
Now, divide the above equation by -24 into both sides. The inequality reverses on multiplying both sides by -1.
-24x/-24 ≥ -96/-24
x ≤ 4

Therefore, the answer is x ≤ 4.

(i) Given inequation 6 + \(\frac { 10x }{ 3 } \) < 4x + 7, where x ∈ Z.
-24x ≥ -96
Solve the inequation 6 + \(\frac { 10x }{ 3 } \) < 4x + 7
\(\frac { 18 + 10x }{ 3 } \) < 4x + 7
Now, multiply the above equation by 3 on both sides.
(18 + 10x) < 12x + 21
Move the x terms on one side.
18 < 12x – 10x + 21
18 < 2x + 21
Now, subtract 21 from both sides of the above equation.
18 – 21 < 2x + 21 – 21
-3 < 2x
Now, divide the above equation with 2 into both sides.
-3/2 < 2x/2
x > -3/2
As x ∈ Z, the answer becomes -1.

(ii) Given inequation is 2x – 1 ≤ \(\frac { 18 – x }{ 2 } \), where x ∈ R
Solve the inequation 2x – 1 ≤ \(\frac { 18 – x }{ 2 } \)
4x – 2 ≤ 18 – x
Move the x terms on one side.
4x + x – 2 ≤ 18
5x -2 ≤ 18
Now, add 2 on both sides of the above equation.
5x -2 + 2 ≤ 18 + 2
5x ≤ 20
Now, divide the above equation with 5 into both sides.
5x/5 ≤ 20/5
x ≤ 4

Therefore, the general value of x for which 2x – 1 ≤ \(\frac { 18 – x }{ 2 } \), where x ∈ R is x = 4.