{"id":44261,"date":"2022-05-30T22:30:36","date_gmt":"2022-05-30T17:00:36","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44261"},"modified":"2023-01-25T11:36:51","modified_gmt":"2023-01-25T06:06:51","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-12-ex-12-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-12-ex-12-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1"},"content":{"rendered":"
Solve the following equations (1 to 12):<\/strong> (ii) 3x – 7 = 3(5 – x) Question 2. (ii) 3(2p – 1) = 5 – (3p – 2) Question 3. (ii) 0.3 (6 – x) = 0.4 (x + 8) Question 4. Question 5. Question 6. Question 7. (ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15 Question 8. (ii) \\(\\frac{5 p+2}{8-2 p}=\\frac{7}{6}\\) Question 9. Question 10. Question 11. (ii) \\(\\frac{2 y+3}{3 y+2}=\\frac{4 y+5}{6 y+7}\\) Question 12. Question 13. ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 12 Linear Equations and Inequalities in one Variable Ex 12.1 Solve the following equations (1 to 12): Question 1. (i) 5x – 3 = 3x – 5 (ii) 3x – 7 = 3(5 – x) Solution: (i) 5x – 3 = 3x – 5 \u21d2 5x … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nQuestion 1.
\n(i) 5x – 3 = 3x – 5
\n(ii) 3x – 7 = 3(5 – x)
\nSolution:
\n(i) 5x – 3 = 3x – 5
\n\u21d2 5x – 3x = -5 + 3
\n\u21d2 2x = \u2014 2
\n\u21d2 x = \\(\\frac{-2}{2}\\) = -1<\/p>\n
\n\u21d2 3x – 7 = 15 – 3x
\n\u21d2 3x + 3x = 15 + 7
\n\u21d2 6x = 22
\n\u21d2 x = \\(\\frac{22}{6}=\\frac{11}{3}\\)<\/p>\n
\n(i) 4(2x + 1) = 3(x – 1) + 7
\n(ii) 3(2p – 1) = 5 – (3p – 2)
\nSolution:
\n(i) 4(2x + 1) = 3(x – 1) + 7
\n\u21d2 8x + 4 = 3x – 3 + 7
\n\u21d2 8x + 4 = 3x + 4
\n\u21d2 8x – 3x = 0
\n\u21d2 x = 0<\/p>\n
\n\u21d2 6p – 3 = 5 – 3p + 2
\n\u21d2 6p + 3p = 5 + 3 + 2
\n\u21d2 9p = 10
\n\u21d2 p = \\(\\frac{10}{9}=1 \\frac{1}{9}\\)<\/p>\n
\n(i) 5y – 2[y – 3(y – 5)] = 6
\n(ii) 0.3(6 – x) = 0.4(x + 8)
\nSolution:
\n(i) 5y – 2[y – 3(y – 5)] = 6
\n\u21d2 5y – 2[y – 3y + 15] = 6
\n\u21d2 5y – 2[-2y + 15] = 6
\n\u21d2 5y + 4y – 30 = 6
\n\u21d2 9y = 6 + 30
\n\u21d2 9y = 36
\n\u21d2 y = \\(\\frac{36}{9}\\)
\n\u21d2 y = 4<\/p>\n
\n\u21d2 1.8 – 0.3x = 0.4x + 3.2
\n\u21d2 0.3x – 0.4x = 3.2 – 1.8
\n\u21d2 -0.7x = 1.4
\n\u21d2 x = \\(\\frac{-1 \\cdot 4}{0 \\cdot 7}\\)
\n\u21d2 x = \\(\\frac{-14}{7}\\)
\n\u21d2 x = -2<\/p>\n
\n(i) \\(\\frac{x-1}{3}=\\frac{x+2}{6}+3\\)
\n(ii) \\(\\frac{x+7}{3}=1+\\frac{3 x-2}{5}\\)
\nSolution:
\n
\n\u21d2 x – 4 = 6 \u00d7 3
\n\u21d2 x – 4 = 18
\n\u21d2 x = 18 + 4
\n\u21d2 x = 22<\/p>\n
\n\u21d2 -4x + 41 = 15
\n\u21d2 -4x = 15 – 41
\n\u21d2 -4x = -26
\n\u21d2 x = \\(\\frac{-26}{-4}\\)
\n\u21d2 x = \\(\\frac{13}{2}\\)
\n\u21d2 x = \\(6 \\frac{1}{2}\\)<\/p>\n
\n
\nSolution:
\n
\n\u21d2 3 (-y + 5) = 6 (1 + 2y)
\n\u21d2 -3y +15 = 6+ 12y
\n\u21d2 -3y – 12y = 6 – 15
\n\u21d2 -15y = -9
\n
\n\u21d2 47p + 480 = 55 \u00d7 60
\n\u21d2 47p + 480 = 3300
\n\u21d2 47p = 3300 – 480
\n\u21d2 47p = 2820
\n\u21d2 p = \\(\\frac{2820}{47}\\)
\n\u21d2 p = 60<\/p>\n
\n
\nSolution:
\n
\n\u21d2 3 (n + 1) = 2(5 – n)
\n\u21d2 3n + 3 = 10 – 2n
\n\u21d2 3n + 2n = 10 – 3
\n\u21d2 5n = 7 \u21d2 n = \\(\\frac{7}{5}=1 \\frac{2}{5}\\)
\n
\n\u21d2 (6t – 4) + (6t + 9) = 6t + 7
\n\u21d2 12t + 5 = 6t + 7
\n\u21d2 12t – 6t = 7 – 5
\n\u21d2 6t = 2 \u21d2 t = \\(\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n
\n(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
\n(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15
\nSolution:
\n(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
\n\u21d2 12x + 8 – 30x + 5 = 2x – 16 – 42x + 24
\n\u21d2 -18x + 13 = -40x + 8
\n\u21d2 -18x + 40x = 8 – 13
\n\u21d2 22x = -5 \u21d2 x = \\(\\frac{-5}{22}\\)<\/p>\n
\n\u21d2 15x + 21 + 10x – 55 = 24x – 15 – 15
\n\u21d2 25x – 34 = 24x – 30
\n\u21d2 25x – 24x = -30 + 34
\n\u21d2 x = 4<\/p>\n
\n
\nSolution:
\n(i) \\(\\frac{3-2 x}{2 x+5}=-\\frac{3}{11}\\)
\n\u21d2 11(3 – 2x) = -3(2x + 5)
\n\u21d2 33 – 22x = -6x – 15
\n\u21d2 -22x + 6x = -15 – 33
\n\u21d2 -16x = -48
\n\u21d2 x = \\(\\frac{48}{16}\\) = 3<\/p>\n
\n\u21d2 6(5p + 2) = 7(8 – 2)p
\n\u21d2 30p + 12 = 56 – 14p
\n\u21d2 30p +14p = 56 – 12
\n\u21d2 44p = 44
\n\u21d2 p =\\(\\frac{44}{44}\\) = 1<\/p>\n
\n
\nSolution:
\n
\n5 (x – 4) = 7x
\n\u21d2 5x – 20 = 7x
\n\u21d2 5x – 7x = 20
\n\u21d2 -2x = 20
\n
\n\u21d2 4(x + 4) = 5(2x + 3)
\n\u21d2 4x + 16 = 10x + 15
\n\u21d2 -6x = -1
\n\u21d2 x = \\(\\frac{-1}{-6}=\\frac{1}{6}\\)<\/p>\n
\n
\nSolution:
\n
\n\u21d2 2x2<\/sup> – 7x – 5 = x (2x – 2)
\n\u21d2 2x2<\/sup> – 7x – 5 = 2x2<\/sup> – 2x
\n\u21d2 -7x – 5 = -2x
\n\u21d2 -7x + 2x = 5
\n\u21d2 -5x = 5
\n\u21d2 x = \\(\\frac{5}{-5}\\)
\n\u21d2 x = \\(\\frac{-5}{5}\\)
\n\u21d2 x = -1
\n
\n\u21d2 3x(1 – 15x) = 15x (3x – 1)
\n\u21d2 3(1 – 15x) = 15 (3x – 1)
\n\u21d2 3 – 45x = 45x – 15
\n\u21d2 -45x – 45x = -15 – 3
\n\u21d2 -90x = -18
\n\u21d2 x = \\(\\frac{-18}{-90} \\Rightarrow x=\\frac{1}{5}\\)<\/p>\n
\n
\nSolution:
\n
\n\u21d2 (2x – 3)(3x + 1) = (3x – 1) (2x – 1)
\n\u21d2 6x2<\/sup> + 2x – 9x – 3 = 6x2<\/sup> – 3x – 2x + 1
\n\u21d2 6x2<\/sup> – 7x – 3 = 6x2<\/sup> – 5x + 1
\n\u21d2 6x2<\/sup> – 7x – 6x2<\/sup> + 5x = 1 + 3
\n\u21d2 -2x = 4 \u21d2 x = \\(\\frac{4}{-2}\\) = -2<\/p>\n
\n\u21d2 (2y + 3) (6y + 7) = (4y + 5) (3y + 2)
\n\u21d2 12y2<\/sup> + 14y + 18y + 21 = 12y2<\/sup> + 8y + 15y + 10
\n\u21d2 32y + 21 = 23y + 10
\n\u21d2 32y – 23y = 10 – 21
\n\u21d2 9y = -11 \u21d2 y = \\(\\frac{-11}{9}\\)<\/p>\n
\nIf x = p + 1, find the value of p from the equation \\(\\frac{1}{2}\\) (5x – 30) – \\(\\frac{1}{3}\\) (1 + 7p) = \\(\\frac{1}{4}\\)
\nSolution:
\nGiven x =p + 1 …(i)
\nAlso \\(\\frac{1}{2}\\)(5x – 30) – \\(\\frac{1}{3}\\)(1 + 7p) = \\(\\frac{1}{4}\\) …(ii)
\nPutting the value of x from (i) in (ii), we get,
\n
\n\u21d2 4 \u00d7 (p – 77) = 1 \u00d7 6
\n\u21d2 4p – 308 = 6
\n\u21d2 4p = 6 + 308
\n\u21d2 4p = 314
\n<\/p>\n
\nSolve \\(\\frac{x+3}{3}-\\frac{x-2}{2}=1\\), Hence find p if \\(\\frac{1}{x}+p\\) = 1.
\nSolution:
\n<\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"