# Worksheet on Laws of Inequality | Free Printable Math Laws of Inequality Worksheet with Answers PDF

Worksheet on Laws of Inequality will help you to learn inequality equations and which inequality statements are true. Students can crack their exams easily by practicing all the problems available in the Laws of Inequality Worksheets. As we are providing all the 10th grade math worksheets for free, students can download them and practice them at any time. Our worksheets on Laws of Inequality are available online and also offline. Therefore, practice them without missing any problem and score good marks in the exam.

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Check out various problems and also various solving methods of inequality problems on this page. We have provided step by step procedure with a detailed explanation for every problem.

Question 1. State true or false.
(i) m < – n ⟹ -m > n
(ii) -10x ≥ 30 ⟹ 2x ≥ -6
(iii) 4x ≤ -14 ⟹ -x ≥ $$\frac { -7 }{ -2 }$$
(iv) 14 > 10 ⟹ $$\frac { 2 }{ 7 }$$ < $$\frac { 2 }{ 5 }$$

Solution:

(i) Given that m < – n ⟹ -m > n
m < – n
As we know that the inequality reverses on multiplying both sides by -1.
-m > n

Therefore, the given equation m < – n ⟹ -m > n is true.

(ii) Given that -10x ≥ 30 ⟹ 2x ≥ -6
-10x ≥ 30
Divide the above equation with -5 into both sides. The inequality reverses on multiplying both sides by -1.
2x ≤ -6

Therefore, the given equation -10x ≥ 30 ⟹ 2x ≥ -6 is false.

(iii) Given that 4x ≤ -14 ⟹ -x ≥ $$\frac { -7 }{ -2 }$$
4x ≤ -14
Divide the above equation with 4 into both sides.
x ≤ $$\frac { -7 }{ 2 }$$
Now, multiply the above equation with -1 into both sides. The inequality reverses on multiplying both sides by -1.
-x ≥ $$\frac { -7 }{ -2 }$$

Therefore, the given equation 4x ≤ -14 ⟹ -x ≥ $$\frac { -7 }{ -2 }$$ is true.

(iv) Given that 14 > 10 ⟹ $$\frac { 2 }{ 7 }$$ < $$\frac { 2 }{ 5 }$$
14 > 10
Divide the above equation with 2 into both sides.
7 > 5
$$\frac { 2 }{ 7 }$$ < $$\frac { 2 }{ 5 }$$

Therefore, the given equation 14 > 10 ⟹ $$\frac { 2 }{ 7 }$$ < $$\frac { 2 }{ 5 }$$ is true.

Question 2. Find the values of x from the given inequations.
(i) – 2x ≥ -3x + 5, find the least positive value of x.
(ii) 2x + 4 ≤ 4x + 6, find the value of x.
(iii) $$\frac { 4x }{ 3 }$$ ≤ $$\frac { 2x }{ 6 }$$ + 2, find the value of x.
(iv) 4x + 8 ≥ 8x, what is the highest positive number of x?

Solution:

(i) Given that – 2x ≥ -3x + 5
– 2x ≥ -3x + 5
Now, subtract 5 from both sides of the above equation.
– 2x – 5 ≥ -3x – 5 + 5
– 2x – 5 ≥ -3x
Now, move -2x to the right side of the above equation.
-5 ≥ -3x + 2x
-5 ≥ -x
Now, multiply the above equation with -1 on both sides.
5 ≤ x

Therefore, the least positive value of x is 1.

(ii) Given that 2x + 4 ≤ 4x + 6.
2x + 4 ≤ 4x + 6
Now, move the x terms on one side of the above equation.
4 ≤ 4x – 2x + 6
4 ≤ 2x + 6
Now, subtract the above equation from 6 on both sides.
4 – 6 ≤ 2x + 6 – 6
-2 ≤ 2x
Divide the above equation with 2 into both sides.
-1 ≤ x

Therefore, the value of x is -1.

(iii) Given that $$\frac { 4x }{ 3 }$$ ≤ $$\frac { 2x }{ 6 }$$ + 2
$$\frac { 4x }{ 3 }$$ ≤ $$\frac { 2x }{ 6 }$$ + 2
Now, multiply the above equation with 3 on both sides.
3$$\frac { 4x }{ 3 }$$ ≤ 3$$\frac { 2x }{ 6 }$$ + 2 (3)
4x ≤ $$\frac { 2x }{ 2 }$$ + 6
4x ≤ x + 6
Now, move the x to the left side of the above equation.
4x – x ≤ 6
3x ≤ 6
Now, divide the above equation with 3 into both sides.
3x/3 ≤ 6/3
x ≤ 2.

Therefore, the value of x is 2.

(iv) Given that 4x + 8 ≥ 8x
4x + 8 ≥ 8x
Move 4x to the right side of the above equation.
8 ≥ 8x – 4x
8 ≥ 4x
Now, divide the above equation with 4 into both sides.
8/4 ≥ 4x/4
2 ≥ x

Therefore, the highest positive number of x is 2.

Question 3. State, whether the following statements are true or false.
(i) If mx + n > p then mx > p – n
(ii) If mx + n > p then x < $$\frac { p – n }{ m }$$, where m is a negative number.
(iii) If m > n then m + p > n + p
(iv) If m > n then m – p > n – p
(v) If m < n, then m – p < n – p
(vi) If m < n then mp < np, p ≠ 0.
(vii) If m < n, then mp > np
(viii) If m > n then $$\frac { m }{ p }$$ > $$\frac { n }{ p }$$, p ≠ 0.
(ix) If m – p < n – l then m + l < n + p.
(x) If m < n, and p > 0, then m – p > n – p, where m, n, p and l are real numbers and p ≠ 0.

Solution:

(i) Given that if mx + n > p then mx > p – n
Take mx + n > p
Now, subtract n on both sides of the above equation.
mx + n – n > p – n
mx > p – n

Therefore, if mx + n > p then mx > p – n. The given equation is true.

(ii) Given that if mx + n > p then x < $$\frac { p – n }{ m }$$, where m is a negative number.
Take mx + n > p
Now, subtract n on both sides of the above equation.
mx + n – n > p – n
mx > p – n
Now, divide the above equation with m on both sides.
mx/m > (p – n)/m
x > (p – n)/m
Given that m is a negative number. Multiply the above equation with -1 n on both sides. The inequality reverses on multiplying both sides by -1.
x < (p – n)/m

Therefore, if mx + n > p then x < $$\frac { p – n }{ m }$$, where m is a negative number. The given equation is true.

(iii) Given that if m > n then m + p > n + p.
Take m > n
Now, add p on both sides of the equation.
m + p > n + p

Therefore, if m > n then m + p > n + p. The given equation is true.

(iv) Given that if m > n then m – p > n – p
Take m > n
Now, subtract p from both sides of the equation.
m – p > n – p

Therefore, if m > n then m – p > n – p. The given equation is true.

(v) Given that if m < n, then m – p < n – p.
Take m < n
Now, subtract p from both sides of the equation.
m – p < n – p

Therefore, if m < n, then m – p < n – p. The given equation is true.

(vi) Given that if m < n then mp < np, p ≠ 0.
Take m < n
Now, multiply p on both sides of the equation.
mp < np
If p is a negative number (p < 0) or p = -1, then the above equation becomes mp > np.

Therefore, the given equation if m < n then mp < np, p ≠ 0 is not true. The given equation is false.

(vii) Given that if m < n, then mp > np.
Take m < n
Now, multiply p on both sides of the equation.
mp < np.

Therefore, the given equation if m < n then mp > np is not true. The given equation is false.

(viii) Given that if m > n then $$\frac { m }{ p }$$ > $$\frac { n }{ p }$$, p ≠ 0.
Take m > n
Now, divide p on both sides of the equation.
$$\frac { m }{ p }$$ > $$\frac { n }{ p }$$
Given p ≠ 0. If p is a negative number (p < 0) or p = -1, then the above equation becomes $$\frac { m }{ p }$$ < $$\frac { n }{ p }$$.

Therefore, the given equation if m > n then $$\frac { m }{ p }$$ > $$\frac { n }{ p }$$, p ≠ 0. is not true when p < 0. The given equation is false.

(ix) Given that if m – p < n – l then m + l < n + p.
Take m – p < n – l
Now, add l on both sides of the above equation.
m – p + l < n – l + l
m – p + l < n
Now, add p on both sides of the above equation.
m – p + l + p < n + p
m + l < n + p

Therefore, if m – p < n – l then m + l < n + p. The given equation is true.

(x) Given that if m < n, and p > 0, then m – p > n – p, where m, n, p and l are real numbers and p ≠ 0.
Take m < n
Now, subtract p from both sides of the equation.
m – p < n – p
Even if p > 0 and m, n, p, and l are real numbers and p ≠ 0, the given equation will not become m – p > n – p.

Therefore, the given statement if m < n, and p > 0, then m – p > n – p, where m, n, p, and l are real numbers and p ≠ 0 is not true. The given equation is false.

Question 4. If a > -6, find the least positive and negative integral values of a.

Solution:

Given that a > -6
The numbers that are greater than -6 are -5, -4, -3, -2, -1, 0, 1, 2, ….
The least positive integral value of a is 1 and the least integral value of a which is next to -6 is -5.

Least positive number = 1, least negative number = -5.

Question 5. (i) If (a – 3)(a – 6) < 0 and a ∈ N then find a.
(ii) If (a + 4)(3 – a) > 0 and a ∈ Z then find a.

Solution:

(i) Given that (a – 3)(a – 6) < 0 and a ∈ N.
We can write (a – 3)(a – 6) < 0 as (a – 3) > 0; (a – 6) < 0
a – 3 > 0
Now, add 3 on both sides of the above equation.
a – 3 + 3 > 0 + 3
a > 3.
a – 6 < 0
Now, add 6 on both sides of the above equation.
a – 6 + 6 < 0 + 6
a < 6.
a € (3, 6)
a = 4, a = 5

Therefore values of a are 4 and 5.

(ii) Given that (a + 4)(3 – a) > 0 and a ∈ Z.
We can write (a + 4)(3 – a) > 0 as (a + 4) > 0; (3 – a) < 0
a + 4 > 0
Now, subtract 4 from both sides of the above equation.
a + 4 – 4 > 0 – 4
a > – 4.
3 – a < 0
Now, Move -a to the right side of the above equation.
a < 3.
a € (-4, 3)
a = -3, a = -2, a = -1, a = 0, a = 1, and, a = 2.

Therefore values of a are -3, -2, -1, 0, 1, and 2.

Question 6. If 10 – 4m ≥ 2, what is the maximum value of m?

Solution:

Given that 10 – 4m ≥ 2.
Now, subtract 10 from both sides of the above equation.
10 – 10 – 4m ≥ 2 – 10.
-4m ≥ – 8.
Now, divide the above equation -4 on both sides. The inequality reverses on multiplying or dividing both sides by -1.
-4m/-4 ≥ – 8/-4.
m ≤ 2.
m = 2, 1, 0, …..

Therefore, the maximum value of m is 2.

Question 7. If a > b and c < 0, which of the following is correct?
(i) ac > bc
(ii) ac < bc

Solution:

Given that a > b and c < 0.
Take a > b.
Now, multiply c on both sides of the above equation.
ac > bc.
But given that c < 0 that means c is a negative number.
If c = -1, then ac > bc becomes ac < bc.

Therefore, ac < bc is the correct answer. The answer is (ii) ac < bc

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