what is the solution set to the inequality (4x-3)(2x-1)>0 ?

ICSE Course viii Maths Selina Solutions are useful for students to score well in their examination. For the benefit of the students, we have provided detailed solutions of the Chapter 15 Linear Inequations. The solutions comprise elaborate steps roofing all the problems that are mentioned under the exercise. The Selina Solutions have been created by the experts by explaining each and every step. This will as well assist students in fetching more marks in the annual exam if they write the answers as provided in the Selina solution pdf.

Solving these questions mainly help students to clear all the concepts about Linear Inequations. Students tin can download ICSE Class 8 Maths Selina Solutions Chapter xv Linear Inequations in pdf format from the link provided  beneath. They can also utilize this pdf for reference during revision. Click on the link beneath to download the ICSE Class 8 Maths Selina solutions of Linear Inequations.

Download ICSE Class viii Maths Selina Solutions Chapter 15:-Download Hither

Affiliate fifteen of ICSE Class 8 Maths Selina Solutions deals with the Linear Inequations. It has one do that contains a total of 15 questions. The detailed pace by stride solutions of the questions has been mentioned below.

CHAPTER fifteen – LINEAR INEQUATIONS

Exercise

Question 1.

If the replacement set is the set of natural numbers, solve.

(i) 10 – five < 0

Solution:-

x – five < 0

Adding v, x – 5 + 5 < 0 + 5…

x < 5

Required answer = {1, 2, iii, 4}

(2) x + 1 < seven

Solution:-

Subtracting 1x + 1 ≤ 7 ⇒ x + ane – 1 ≤ 7 – 1

x ≤ 6

Required respond = {one, 2, 3, 4, 5, half-dozen}

(iii) 3x – four > half dozen

Solution:-

3x – iv > 6

Adding four, 3x – 4 + iv > 6 + four

3x > 10

Required respond = {4, 5, 6,…}

(iv) 4x + 1 > 17

Solution:-

4x + 1 ≥ 17

Subtracting, 4x + i – 1 ≥ 17 – one

4x ≥ sixteen

Dividing by 4, 4x/4 ≥ (16/4)10 ≥ 4

Required answer = {4, v, 6,…}

Question 2.

If the replacement set = {-6, -3, 0, three, 6, 9}; find the truth set of the following:

(i) 2x – 1 > 9

Solution:-

2x – 1 > 9

Adding one, 2x – 1 + 1 > 9 + 1

2x >x

Dividing past ii, x > 5

Required answer = {6, 9}

(ii) 3x + 7 < ane

Solution:-

3x + 7 ≤ 1

Subtracting vii, 3x + 7 – seven ≤ 1 – 7

3x ≤ -6

x ≤ -two

Required Reply = {-vi, -3}

Question 3.

Solve 7 > 3x -8; x ∈ Due north

Solution:-

7 > 3x – viii

Subtracting 3x, 7 – 3x > 3x – 3x – 8

Subtracting 7, 7 – 7 – 3x > 3x – 3x – 8 – vii

-3x > -15

Dividing by -3, x < 5

Required Answer = {1, two, 3, 4}

Notation: Division by negative number reverses the inequality

Question four

-17 < 9y – viii; y ∈ Z

Solution:-

-17 < 9y – 8

Adding eight, – 17 + 8 < 9y – viii + 8

– ix < 9y

Dividing by 9

-1 < y

Required Answer = {0, ane, two, 3, 4 …. }

Question 5.

Solve 9x – seven ≤ 28 + 4x; x ∈ W

Solution:-

9x – 1 ≤ 28 + 4x

Subtracting 4x, 9x – 4x – 7 ≤ 28 + 4x – 4x

5x – vii ≤ 28

Calculation seven, 5x – vii + seven ≤ 28 + seven

5x ≤ 35

Dividing by 5, x≤7

Required answer = {0, ane, 2, 3, four, 5, 6, 7}

Question 6.

Solve ii/3x + 8 < 12; ten ∈ Due west

Solution:-

Multiplying by three/2, (ii/three) x × (3/2) < four × (3/2)

∴ Required answer = {0, ane, ii, 3, iv, 5}

Question 7.

Solve − 5(x + 4) > 30; x ∈ Z

Solution:-

-5(x + 4) > 30

Dividing by -v, ((−5(x+4))/−5) < (30/−five)

Note: Division by a negative number reverses the equality

x + four < -6

x + iv – 4 < – half dozen – 4

x < – ten

∴ Required Answer = {-xi, -12, -13, …}

Question 8.

Solve the inequation eight – 2x > x – five; x ∈ N

Solution:-

8 – 2x ≥ 10 – 5; 10 ∈ North

8 + 5 ≥ 2x + 10

thirteen ≥ 3x ⇒ 3x ≤ 13

x = ane, 2, 3, four (x ∈ N)

Solution set = {1, 2, 3, 4}

Question 9.

Solve the inequality xviii -3 (2x – 5) > 12; 10 ∈ W.

Solution:-

18 – three(2x – 5) > 12; 10 ∈ Westward

18 – 6x + 15 > 12

33 – 12 > 6x

21 > 6x

6x < 21 ⇒ x < 21/6 + seven/ii =3½

But x ∈ W, x = 0, 1, 2, three

∴ Solution set = {0, i, two, three}

Question ten.

Solve: ((2x+1)/3) + 15 < 17; x ∈ W

Solution:-

((2x+1)/3) + 15 ≤ 17; ten ∈ Westward ((2x+i)/3) ≤ 17 – 15 = ii

2x + 1 ≤ 6 ⇒ 2x ≤ 5

10 ≤ 5/2 = 2½

∴ 10 = 0, 1, 2

∴ Solution set is = {0, 1, 2}

Question 11.

Solve:- 3 + 10 < 2, x ∈ N

Solution:

-three + x < 2, x ∈ Due north

x < 2 – (-3)

10 < 2 + 3

10 < 5

∴ ten = 1, 2, three, 4 (∵x ∈ N)

∴ Solution set = {1, two, 3, 4}

Question 12.

Solve: 4x – v > x – ten, x ∈ {0, 1, 2, iii, iv, 5, six, 7}

Solution:

4x – 5 > 10 – x, ten ∈ N

4x + x > 10 + 5

5x > 15

10 > 15/v = three

∴10=4, 5, 6, 7

Solution set = {4, 5, vi, 7}

Question thirteen.

Solve: 15 – 2(2x – i) < 15, x ∈ Z

Solution:

15 – 4x + 2 < 15

17 – 4x < 15

-4x < fifteen – 17

-4x < -2

Dividing by -four, (−4/−iv)x > −2/−4 = ½

∴x = 1, two, 3, 4, v,

∴ Solution ready = {1, two, three, 4, five,…}

Question xiv.

Solve: (2x + 3)/5 > (4x−1)/2, x ∈ W

Solution:-

(2x + 3)/v > (4x − ane)/ii, x ∈ Due west

2(2x + 3) > v(4x – 1)

4x + half-dozen > 20x – 5

4x – 20x > – five – 6

-16x > -xi

Dividing past -16, x< (−eleven/−16) x < (11/sixteen)

∴ x = 0

∴ Solution set = {0}

Solve and graph the solution assault a number line:

Question 15.

x – 5 < – 2; x ∈ N

Solution:-

ten – 5 < – 2

Adding five to both sides, x – 5 + 5 < – 2 + 5

ten < 3

∴ The required graph is

ICSE Class 8 Maths Selina Solutions Chapter xv – Linear Inequations

Almost everyday nosotros come across mathematical inequalities, only you might not have noticed them considering they are so familiar. Some of the examples of mathematical inequalities that nosotros come across in our daily lives are speed limit in highways, minimum payments on credit card bills, etc. While considering these situations nosotros employ mathematical thinking such as speed limits on highway can be mathematically written every bit Legal speed on the highway ≤ 65 miles per 60 minutes. When two expressions are connected past "greater than" or "less than" sign, nosotros get an inequality. Also, the linear inequation is similar to a linear equation, where the equal to sign is replaced by the inequality sign.

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