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

Take the help of the Worksheet on Solution of a Linear Inequation in One Variable while preparing Linear Inequation. We have provided the difference between the Linear equation and Linear Inequation along with Linear Inequation Worksheet. Check out various problems we have given on the worksheet and prepare well for the exam.

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## Solving Linear Inequalities in One Variable Worksheet Answer Key

Check out the problems available on One Variable Linear Inequations and practice all the problems without missing anyone.

Question 1. If x ∈ N, find the solution set of the linear inequations.
(i) 10x + 6 ≤ 4x + 36
(ii) 6x – 4 < 38 – 8x

Solution:

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}.

Question 2. (i) Is x = -4 a solution of inequation 8x + 6 < 6x – 2? Why?
(ii) Is x = 2 a solution of inequation 4x + 2 ≥ 2x – 6? Why?

Solution:

(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.

Question 3. Solve the inequation: 9 – 6x ≥ 3x – 36 given that x ∈ N.

Solution:

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}.

Question 4. Solve the inequations in R:
(i) 2x – 4 > 6
(ii) 6x < 30
(iii) -12x ≥ -48
(iv) 16x – 12 ≥ 36
(v) 10 – 4x < 230.

Solution:

(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.

Question 5. If 100 – 16x ≤ 64, find:
(i) the smallest value of x, when x is a real number,
(ii) The smallest value of x, when x is an integer.

Solution:

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.

Question 6. x is a positive integer satisfying 90 – 12(6x + 3) < 90. Find the solution set of the inequation.

Solution:

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, …..}

Question 7. Solve the inequations in R:
(i) -2x + 14 > 8x – 6
(ii) 14x – 10x ≥ 6 + 2x
(iii) 4(2x + 1) ≤ x + 10
(iv) 10(6x – 4) < 6(8x – 6)
(v) 6 + $$\frac { x }{ 2 }$$ > $$\frac { 2x }{ 5 }$$ + 14
(vi) $$\frac { 2x – 1 }{ 7 }$$ ≥ $$\frac { 2x + 3 }{ 3 }$$

Solution:

(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.

Question 8. If x and y be positive integers satisfying x + y ≤ 4. What are the possible values of x and y?

Solution:

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)

Question 9. Find the largest value of x for which 4(x – 1) ≤ 18 – 2x and x ∈ W.

Solution:

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.

Question 10. Solve the inequations:
(i) 6 + 10x > 6x – 6, where x is a negative integer
(ii) 10x + 8 < 6x + 40, where x ∈ N.
(iii) x + 5 ≤ $$\frac { 2x }{ 3 }$$ + 6, where x is a positive odd integer.
(iv) 4x + 6 ≥ 2x + 10, where x is a natural number less than 8.
(v) $$\frac { 2x + 6 }{ 3 }$$ ≤ $$\frac { 2x + 16 }{ 4 }$$, where x is positive even integer.
(vi) $$\frac { 6x }{ 5 }$$ – $$\frac { 4 (x – 2) }{ 3 }$$ > 2, where x ∈ {2, 4, 6, 8, 10}

Solution:

(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}

Question 11. Solve the inequation: -24x ≥ -96

Solution:

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.

Question 12. (i) Find the smallest value of x for which 6 + $$\frac { 10x }{ 3 }$$ < 4x + 7, where x ∈ Z.
(ii) Find the general value of x for which 2x – 1 ≤ $$\frac { 18 – x }{ 2 }$$, where x ∈ R

Solution:

(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.

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