Solve the following linear equations.<\/span><\/p>\nQuestion 1. \n\\(\\frac{x}{2}-\\frac{1}{5}=\\frac{x}{3}+\\frac{1}{4}\\) \nSolution: \n\\(\\frac{x}{2}-\\frac{1}{5}=\\frac{x}{3}+\\frac{1}{4}\\) \nTransposing \\(-\\frac{1}{5}\\) to R.H.S. and \\(\\frac{\\mathrm{x}}{3}\\) to L.H.S \n\\(\\frac{x}{2}-\\frac{x}{3}=\\frac{1}{4}+\\frac{1}{5}\\) \n\\(\\frac{3 x-2 x}{6}=\\frac{5+4}{20}\\) \n\\(\\frac{x}{6}=\\frac{9}{20}\\) \nBy cross-multiplication, we get \n20(x) = 9 \u00d7 6 \n20x = 54 \nDividing both sides by 20 \n\\(\\frac{20 \\mathrm{x}}{20}=\\frac{54}{20}\\) \nx = \\(\\frac{27}{10}\\)<\/p>\n
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Question 2. \n\\(\\frac{\\mathrm{n}}{2}-\\frac{3 n}{4}+\\frac{5 \\mathrm{n}}{6}=21\\) \nSolution: \n\\(\\frac{\\mathrm{n}}{2}-\\frac{3 n}{4}+\\frac{5 \\mathrm{n}}{6}=21\\) \nL.C.M of 2, 4 and 6 = 12 \nMultiplying each term by 12, we get \n\\(12 \\times \\frac{\\mathrm{n}}{2}-12 \\times \\frac{3 \\mathrm{n}}{4}+12 \\times \\frac{5 \\mathrm{n}}{6}=21 \\times 12\\) \n6n – 9n + 10n = 252 \n7n = 252 \nDividing both sides by 7 \n\\(\\frac{7 \\mathrm{n}}{7}=\\frac{252}{7}\\) \nn = 36<\/p>\n
Question 3. \n\\(x+7-\\frac{8 x}{3}=\\frac{17}{6}-\\frac{5 x}{2}\\) \nSolution: \n\\(x+7-\\frac{8 x}{3}=\\frac{17}{6}-\\frac{5 x}{2}\\) \nTransposing 7 to R.H.S. and \\(\\frac{-5 x}{2}\\) to L.H.S \n\\(x-\\frac{8 x}{3}+\\frac{5 x}{2}=\\frac{17}{6}-7\\) \nL.C.M of 3, 2 and 6 = 6 \nMultiplying each term by 6, we get \n\\(6 \\times x-6 \\times \\frac{8 x}{3}+6 \\times \\frac{5 x}{2}=6 \\times \\frac{17}{6}-6 \\times 7\\) \n6x – 16x + 15x = 17 – 42 \n5x = -25 \nDividing both sides by 5 \n\\(\\frac{5 x}{5}=-\\frac{25}{5}\\) \nx = -5<\/p>\n
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Question 4. \n\\(\\frac{x-5}{3}=\\frac{x-3}{5}\\) \nSolution: \n\\(\\frac{x-5}{3}=\\frac{x-3}{5}\\) \nBy cross-multiplication we get \n5(x – 5) = 3(x – 3) \n5x – 25 = 3x – 9 \nTransposing -25 to R.H.S and 3x to L.H.S. \n5x – 3x = 25 – 9 \n2x = 16 \nDividing both sides by 2 \n\\(\\frac{2 \\mathrm{x}}{2}\\) = \\(\\frac{16}{2}\\) \nx = 8<\/p>\n
Question 5. \n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}=\\frac{2}{3}-\\mathrm{t}\\) \nSolution: \n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}=\\frac{2}{3}-\\mathrm{t}\\) \nTransposing -t to L.H.S., we get \n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}+\\mathrm{t}=\\frac{2}{3}\\) \nL.C.M of 4 and 3 is 12 \nMultiplying each term by 12 \n\\(12 \\frac{(3 \\mathrm{t}-2)}{4}-12 \\frac{(2 \\mathrm{t}+3)}{3}+12 \\mathrm{t}=12 \\times \\frac{2}{3}\\) \n3(3t – 2) – 4(2t + 3) + 12t = 8 \n9t – 6 – 8t – 12 + 12t = 8 \n13t – 18 = 8 \nTransposing -18 to R.H.S., we get \n13t = 8 + 18 \n13t = 26 \nDividing both sides by 13 \n\\(\\frac{13 t}{13}=\\frac{26}{13}\\) \nt = 2<\/p>\n
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Question 6. \n\\(m-\\frac{m-1}{2}=1-\\frac{m-2}{3}\\) \nSolution: \n\\(m-\\frac{m-1}{2}=1-\\frac{m-2}{3}\\) \nTransposing \\(\\left(\\frac{m-2}{3}\\right)\\) to L.H.S. we get \n\\(\\mathrm{m}-\\frac{\\mathrm{m}-1}{2}+\\frac{\\mathrm{m}-2}{3}=1\\) \nL.C.M of 2 and 3 is 6 \nMultiplying each term by 6, we get \n6m – \\(\\frac{6(m-1)}{2}+\\frac{6(m-2)}{3}\\) = 1 \u00d7 6 \n6m – 3(m – 1) + 2(m – 2) = 6 \n6m – 3m + 3 + 2m – 4 = 6 \n5m – 1 = 6 \nTransposing -1 to the R.H.S., we get \n5m = 6 + 1 \n5m = 7 \nDividing both sides by 5, we get \n\\(\\frac{5 m}{5}=\\frac{7}{5}\\) \nm = \\(\\frac{7}{5}\\)<\/p>\n
Simplify and solve the following linear equations.<\/span><\/p>\nQuestion 7. \n3(t – 3) = 5 (2t + 1) \nSolution: \n3(t – 3) = 5 (2t + 1) \nOpening the brackets, we get \n3t – 9 = 10t + 5 \nTransposing -9 to R.H.S. and 10t to L.H.S. \n3t – 10t = 5 + 9 \n-7t = 14 \nDividing both sides by -7, we get \n\\(\\frac{-7 t}{-7}=\\frac{14}{-7}\\) \nt = -2<\/p>\n
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Question 8. \n15(y – 4) – 2(y – 9) + 5 (y + 6) = 0 \nSolution: \n15(y – 4) – 2(y – 9) + 5(y + 6) = 0 \nOpening the brackets, we get \n15y – 60 – 2y + 18 + 5y + 30 = 0 \n15y – 2y + 5y – 60 + 18 + 30 = 0 \n18y – 12 = 0 \nTransposing -12 to R.H.S. \n18y = 12 \nDividing both sides by 18, we get. \n\\(\\frac{18 y}{18}=\\frac{12}{18}\\) \ny = \\(\\frac{2}{3}\\)<\/p>\n
Question 9. \n3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 \nSolution: \n3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 \nOpening the brackets, we get \n15z – 21 – 18z + 22 = 32z – 52 – 17 \n15z – 18z + 22 – 21 = 32z – 69 \n-3z + 1 = 32z – 69 \nTransposing 1 to R.H.S. and 32z to L.H.S. \n-3z – 32z = – 69 – 1 \n-35z = -70 \nDividing both sides by -35, we get \n\\(\\frac{-35 z}{-35}=\\frac{-70}{-35}\\) \nz = 2<\/p>\n
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Question 10. \n0.25(4f – 3) = 0.05(10f – 9) \nSolution: \n0.25(4f – 3) = 0.05(10f – 9) \nOpening the brackets, we get \n0.25(4f) – 0.25 \u00d7 3 = 0.05 \u00d7 10f – 0.05 \u00d7 9 \nf – 0.75 = 0.5f – 0.45 \nf – 0.5f = 0.75 – 0.45 \n0.5f = 0.3 \nDividing both sides by 0.5, we get \n\\(\\frac{0.5 \\mathrm{f}}{0.5}=\\frac{0.3}{0.5}\\) \nf = \\(\\frac{3}{5}\\) = 0.6 \nf = 0.6<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 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 2.5 Solve the following linear equations. Question 1. Solution: Transposing to R.H.S. and to …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5<\/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 2.5 - 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