Solve the following equations.<\/span><\/p>\nQuestion 1. \nx – 2 = 7 \nSolution: \nx – 2 = 7 \nTransposing (-2) to R.H.S., we get \nx = 7 + 2 \n\u2234 x = 9<\/p>\n
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Question 2. \ny + 3 = 10 \nSolution: \ny + 3 = 10 \nTransposing 3 to R.H.S., we get \ny = 10 – 3 \n\u2234 y = 7<\/p>\n
Question 3. \n6 = z + 2 \nSolution: \n6 = z + 2 \nTransposing 2 to L.H.S., we get \n6 – 2 = z \n4 = z \n\u2234 z = 4<\/p>\n
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Question 4. \n\\(\\frac{3}{7}+x=\\frac{17}{7}\\) \nSolution: \n\\(\\frac{3}{7}+x=\\frac{17}{7}\\) \nTransposing \\(\\frac{3}{7}\\) to R.H.S., we get \nx = \\(\\frac{17}{7}-\\frac{3}{7}=\\frac{17-3}{7}=\\frac{14}{7}=2\\) \n\u2234 x = 2<\/p>\n
Question 5. \n6x = 12 \nSolution: \n6x = 12 \nDivided by 6 on both sides, we get \n\\(\\frac{6 x}{6}=\\frac{12}{6}\\) \n\u2234 x = 2<\/p>\n
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Question 6. \n\\(\\frac{t}{5}=10\\) \nSolution: \n\\(\\frac{\\mathrm{t}}{5}\\) = 10 \nMultiplying both sides by 5, we get \n\\(\\frac{\\mathrm{t}}{5}\\) \u00d7 5 = 10 \u00d7 5 \n\u2234 t = 50<\/p>\n
Question 7. \n\\(\\frac{2 x}{3}=18\\) \nSolution: \n\\(\\frac{2 x}{3}=18\\) \nMultiplying both sides, by 3, we get \n\\(\\frac{2 x}{3}\\) \u00d7 3 = 18 \u00d7 3 \n2x = 18 \u00d7 3 \nx = \\(\\frac{18 \\times 3}{2}\\) = 9 \u00d7 3 = 27 \n\u2234 x = 27<\/p>\n
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Question 8. \n1.6 = \\(\\frac{\\mathrm{y}}{1.5}\\) \nSolution: \n1.6 = \\(\\frac{\\mathrm{y}}{1.5}\\) \nMultiplying both sides by 1.5, we get \n1.6 \u00d7 1.5 = \\(\\frac{\\mathrm{y}}{1.5}\\) \u00d7 1.5 \n2.4 = y \n\u2234 y = 2.4<\/p>\n
Question 9. \n7x – 9 = 16 \nSolution: \n7x – 9 = 16 \nTransposing (-9) to R.H.S., we get \n7x = 16 + 9 \nDividing both sides by 7, we get \n\u2234 x = \\(\\frac{25}{7}\\)<\/p>\n
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Question 10. \n14y – 8 = 13 \nSolution: \n14y – 8 = 13 \nTransposing (-8) to R.H.S., we get \n14y = 13 + 8 \n14y = 21 \nDividing both sides by 14 we get, \n\\(\\frac{14 y}{14}\\) = \\(\\frac{21}{14}\\) \n\u2234 y = \\(\\frac{3}{2}\\)<\/p>\n
Question 11. \n17 + 6p = 9 \nSolution: \n17 + 6P = 9 \nTransposing 17 to RHS, we get \n6P = 9 – 17 \n6P = -8 \nDividing both sides by 6, we have \n\\(\\frac{6 P}{6}=\\frac{-8}{6}\\) \n\u2234 P = \\(\\frac{-4}{3}\\)<\/p>\n
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Question 12. \n\\(\\frac{x}{3}+1=\\frac{7}{15}\\) \nSolution: \n\\(\\frac{x}{3}+1=\\frac{7}{15}\\) \nTransposing 1 to R.H.S, we get \n\\(\\frac{x}{3}=\\frac{7}{15}-1\\) \n\\(\\frac{x}{3}=\\frac{7-15}{15}\\) \n\\(\\frac{x}{3}=\\frac{-8}{15}\\) \nMultiplying both sides by 3, we have \n\\(\\frac{x}{3} \\times 3=\\frac{-8}{15} \\times 3\\) \n\u2234 x = \\(\\frac{-8}{5}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1 Solve the following equations. Question 1. x – 2 = 7 Solution: x …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n