ML Aggarwal Class-8 Linear Equations and Inequalities ICSE Maths

ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 Linear Equations and Inequalities in One Variable Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-12.1 , Exe-12.2, Exe-12.3  Objective Type Questions (including Mental Maths Multiple Choice Questions Value Based Questions , HOTS),  and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions Chapter-12


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Exercise 12.1 ,

Exercise-12.2,

Exercise-12.3,

Objective Type Questions, 

Mental Maths,

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Ex 12.1 , ML Aggarwal Class-8 Linear Equations and Inequalities in One Variable ICSE Mathematics Solutions

Solve the following equations (1 to 12):

Question 1.
(i) 5x – 3 = 3x – 5
(ii) 3x – 7 = 3(5 – x)

Answer

(i) 5x – 3 = 3x – 5
⇒ 5x – 3x = -5 + 3
⇒ 2x = — 2
⇒ x = \frac{-2}{2} = -1

(ii) 3x – 7 = 3(5 – x)
⇒ 3x – 7 = 15 – 3x
⇒ 3x + 3x = 15 + 7
⇒ 6x = 22
⇒ x = \frac{22}{6}=\frac{11}{3}

Question 2.
(i) 4(2x + 1) = 3(x – 1) + 7
(ii) 3(2p – 1) = 5 – (3p – 2)

Answer

(i) 4(2x + 1) = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 0
⇒ x = 0

(ii) 3(2p – 1) = 5 – (3p – 2)
⇒ 6p – 3 = 5 – 3p + 2
⇒ 6p + 3p = 5 + 3 + 2
⇒ 9p = 10
⇒ p = \frac{10}{9}=1 \frac{1}{9}

Question 3.
(i) 5y – 2[y – 3(y – 5)] = 6
(ii) 0.3(6 – x) = 0.4(x + 8)

Answer

(i) 5y – 2[y – 3(y – 5)] = 6
⇒ 5y – 2[y – 3y + 15] = 6
⇒ 5y – 2[-2y + 15] = 6
⇒ 5y + 4y – 30 = 6
⇒ 9y = 6 + 30
⇒ 9y = 36
⇒ y = \frac{36}{9}
⇒ y = 4

(ii) 0.3 (6 – x) = 0.4 (x + 8)
⇒ 1.8 – 0.3x = 0.4x + 3.2
⇒ 0.3x – 0.4x = 3.2 – 1.8
⇒ -0.7x = 1.4
⇒ x = \frac{-1 \cdot 4}{0 \cdot 7}
⇒ x = \frac{-14}{7}
⇒ x = -2

Question 4.
(i) \frac{x-1}{3}=\frac{x+2}{6}+3
(ii) \frac{x+7}{3}=1+\frac{3 x-2}{5}

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 1

⇒ x – 4 = 6 × 3
⇒ x – 4 = 18
⇒ x = 18 + 4
⇒ x = 22

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 2

⇒ -4x + 41 = 15
⇒ -4x = 15 – 41
⇒ -4x = -26
⇒ x = \frac{-26}{-4}
⇒ x = \frac{13}{2}
⇒ x = 6 \frac{1}{2}

Question 5.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 3

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 4
⇒ 3 (-y + 5) = 6 (1 + 2y)
⇒ -3y +15 = 6+ 12y
⇒ -3y – 12y = 6 – 15
⇒ -15y = -9
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 5
⇒ 47p + 480 = 55 × 60
⇒ 47p + 480 = 3300
⇒ 47p = 3300 – 480
⇒ 47p = 2820
⇒ p = \frac{2820}{47}
⇒ p = 60

Question 6.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 6

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 7
⇒ 3 (n + 1) = 2(5 – n)
⇒ 3n + 3 = 10 – 2n
⇒ 3n + 2n = 10 – 3
⇒ 5n = 7 ⇒ n = \frac{7}{5}=1 \frac{2}{5}
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 8
⇒ (6t – 4) + (6t + 9) = 6t + 7
⇒ 12t + 5 = 6t + 7
⇒ 12t – 6t = 7 – 5
⇒ 6t = 2 ⇒ t = \frac{2}{6}=\frac{1}{3}

Question 7.
(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15

Answer

(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
⇒ 12x + 8 – 30x + 5 = 2x – 16 – 42x + 24
⇒ -18x + 13 = -40x + 8
⇒ -18x + 40x = 8 – 13
⇒ 22x = -5 ⇒ x = \frac{-5}{22}

(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15
⇒ 15x + 21 + 10x – 55 = 24x – 15 – 15
⇒ 25x – 34 = 24x – 30
⇒ 25x – 24x = -30 + 34
⇒ x = 4

Question 8.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 9

Answer

(i) \frac{3-2 x}{2 x+5}=-\frac{3}{11}
⇒ 11(3 – 2x) = -3(2x + 5)
⇒ 33 – 22x = -6x – 15
⇒ -22x + 6x = -15 – 33
⇒ -16x = -48
⇒ x = \frac{48}{16} = 3

(ii) \frac{5 p+2}{8-2 p}=\frac{7}{6}
⇒ 6(5p + 2) = 7(8 – 2)p
⇒ 30p + 12 = 56 – 14p
⇒ 30p +14p = 56 – 12
⇒ 44p = 44
⇒ p =\frac{44}{44} = 1

Question 9.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 10

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 11
5 (x – 4) = 7x
⇒ 5x – 20 = 7x
⇒ 5x – 7x = 20
⇒ -2x = 20
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 12

⇒ 4(x + 4) = 5(2x + 3)
⇒ 4x + 16 = 10x + 15
⇒ -6x = -1
⇒ x = \frac{-1}{-6}=\frac{1}{6}

Question 10.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 13

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 14
⇒ 2x2 – 7x – 5 = x (2x – 2)
⇒ 2x2 – 7x – 5 = 2x2 – 2x
⇒ -7x – 5 = -2x
⇒ -7x + 2x = 5
⇒ -5x = 5
⇒ x = \frac{5}{-5}
⇒ x = \frac{-5}{5}
⇒ x = -1

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 15

⇒ 3x(1 – 15x) = 15x (3x – 1)
⇒ 3(1 – 15x) = 15 (3x – 1)
⇒ 3 – 45x = 45x – 15
⇒ -45x – 45x = -15 – 3
⇒ -90x = -18
⇒ x = \frac{-18}{-90} \Rightarrow x=\frac{1}{5}

Question 11.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 16

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 17
⇒ (2x – 3)(3x + 1) = (3x – 1) (2x – 1)
⇒ 6x2 + 2x – 9x – 3 = 6x2 – 3x – 2x + 1
⇒ 6x2 – 7x – 3 = 6x2 – 5x + 1
⇒ 6x2 – 7x – 6x2 + 5x = 1 + 3
⇒ -2x = 4 ⇒ x = \frac{4}{-2} = -2

(ii) \frac{2 y+3}{3 y+2}=\frac{4 y+5}{6 y+7}
⇒ (2y + 3) (6y + 7) = (4y + 5) (3y + 2)
⇒ 12y2 + 14y + 18y + 21 = 12y2 + 8y + 15y + 10
⇒ 32y + 21 = 23y + 10
⇒ 32y – 23y = 10 – 21
⇒ 9y = -11 ⇒ y = \frac{-11}{9}

Question 12.

If x = p + 1, find the value of p from the equation \frac{1}{2} (5x – 30) – \frac{1}{3} (1 + 7p) = \frac{1}{4}

Answer

Given x =p + 1 …(i)
Also \frac{1}{2}(5x – 30) – \frac{1}{3}(1 + 7p) = \frac{1}{4} …(ii)
Putting the value of x from (i) in (ii), we get,
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 18
⇒ 4 × (p – 77) = 1 × 6
⇒ 4p – 308 = 6
⇒ 4p = 6 + 308
⇒ 4p = 314
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 19

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