NCERT Solutions for Class 7 Maths<\/a> Chapter 4 Simple Equations Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3<\/h2>\n Question 1. \nSolve the following equations: \n(a) 2y + \\(\\frac{5}{2}=\\frac{37}{2}\\) \n(b) 5t + 28 = 10 \n(c) \\(\\frac{\\mathrm{a}}{5}\\) + 3 = 2 \n(d) \\(\\frac{\\mathrm{q}}{4}\\) + 7 = 5 \n(e) \\(\\frac{5}{2}\\) x = -10 \n(f) \\(\\frac{5}{2}\\) x = \\(\\frac{25}{4}\\) \n(g) 7m + \\(\\frac{19}{2}\\) = 13 \n(h) 6z + 10 = -2 \n(i) \\(\\frac{31}{2}=\\frac{2}{3}\\) \n(j) \\(\\frac{2 \\mathrm{~b}}{3}\\) – 5 = 3 \nAnswer: \n(a) we have \n2y + \\(\\frac{5}{2}=\\frac{37}{2}\\) \nTransposing \\(\\frac{5}{2}\\) from L.H.S to R.H.S \n2y = \\(\\frac{37}{2}-\\frac{5}{2}\\) \n2y = \\(\\frac{37-5}{2}\\) \n2y = \\(\\frac{32}{2}\\) \n2y = 16.<\/p>\n
(b) we have \n5t + 28 = 10 \nTransposing 28 from L.H.S to R.H.S \n5t = 10 – 28 \n5t = – 18 \nDividing both sides by 5, we get \n\\(\\frac{5 \\mathrm{t}}{5}=\\frac{-18}{5}\\) \nt = \\(\\frac{-18}{5}\\) \nThe solution is t = \\(\\frac{-18}{5}\\) or -3 \\(\\frac{3}{5}\\) \nDividing both sides by 2, we get \n\\(\\frac{2 y}{2}=\\frac{16}{2}\\) \ny = 8 \ny = 8 is the required solution.<\/p>\n
(c) we have a \n\\(\\frac{\\mathrm{a}}{5}\\) + 3 = 2 \nTransposing 3 from L.H.S to R.H.S \n\\(\\frac{\\mathrm{a}}{5}\\) = 2 – 3 \n\\(\\frac{\\mathrm{a}}{5}\\) = -1 \nMultiplying both sides by 5, we get \n\\(\\frac{\\mathrm{a} \\times 5}{5}\\) = -1 x 5 \na = – 5 \nThe solution is a = – 5.<\/p>\n
(d) We have \n\\(\\frac{\\mathrm{q}}{4}\\) + 7 = 5 \nTransposing 7 to R.H.S \n\\(\\frac{\\mathrm{q}}{4}\\) = 5 – 7 \n\\(\\frac{\\mathrm{q}}{4}\\) = -2 \nMultiplying both sides by 4, we get \n\\(\\frac{\\mathrm{q}}{4}\\) \u00d7 4 = – 2 \u00d7 (4) \nq = -8 \nThe solution is q = – 8.<\/p>\n
(e) We have \n\\(\\frac{5}{2}\\) x = – 10 \nMultiplying both sides by 2, we get \n\\(\\frac{5}{2}\\) x x 2 = -10 \u00d7 2 \n5x = – 20 \nDividing both sides by 5, we get \n\\(\\frac{5 x}{5}=\\frac{-20}{5}\\) \nThe solution is x = – 4.<\/p>\n
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(f) We have \n\\(\\frac{5}{2}\\) x = \\(\\frac{25}{4}\\) \nMultiplying both sides by 2, we get \n\\(\\frac{5}{2}\\) x \u00d7 2 = \\(\\frac{25}{4}\\) \u00d7 2 \n5x = \\(\\frac{25}{2}\\) \nDividing both sides by 5, we get \n\\(\\frac{5 x}{5}=\\frac{25}{2 \\times 5}\\) \nx = \\(\\frac{5}{2}\\) \nThe solution is x= \\(\\frac{5}{2}\\) or x = 2 \\(\\frac{1}{2}\\).<\/p>\n
(g) We have \n7m + \\(\\frac{19}{2}\\) = 13 \nTransposing \\(\\frac{19}{2}\\) to R.H.S \n7m = 13 – \\(\\frac{19}{2}\\) \n7m = \\(\\frac{26-19}{2}\\) \n7m = \\(\\frac{7}{2}\\)<\/p>\n
(h) We have \n6z + 10 = – 2 \nTransposing 10 to R.H.S \n6z = – 2 – 10 \n6z = – 12 \nDividing both sides by 6, we get \n\\(\\frac{6 z}{6}=\\frac{-12}{6}\\) \nz = – 2 \nThe solution is z = – 2. \nDividing both sides by 7, we get \n\\(\\frac{7 \\mathrm{~m}}{7}=\\frac{7}{2 \\times 7}\\) \nm = \\(\\frac{1}{2}\\) \nThe solution is m = \\(\\frac{1}{2}\\) .<\/p>\n
(i) We have \n\\(\\frac{31}{2}=\\frac{2}{3}\\) \nMultiplying both sides by 2, we get \n\\(\\frac{31}{2}\\) x 2 = \\(\\frac{2}{3}\\) x 2 \n31 = \\(\\frac{4}{3}\\) \nDividing both sides by 3, we get \n\\(\\frac{31}{3}=\\frac{4}{3 \\times 3}\\) \n1 = \\(\\frac{4}{9}\\) \nThe solution is 1 = \\(\\frac{4}{9}\\)<\/p>\n
(j) We have \n\\(\\frac{2 \\mathrm{~b}}{3}\\) – 5 = 3 \nTransposing – 5 to R.H.S \n\\(\\frac{2 \\mathrm{~b}}{3}\\) = 3 + 5 \n\\(\\frac{2 \\mathrm{~b}}{3}\\) = 8 \nMultiplying both sides by 3, we get \n\\(\\frac{2 \\mathrm{~b}}{3}\\) \u00d7 3 = 8 \u00d7 3 \n2b = 24 \nDividing both sides by 2, we get \n\\(\\frac{2 b}{2}=\\frac{24}{2}\\) \nb = 12 \nThe solution is b = 12.<\/p>\n
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Question 2. \nSolve the following equations: \n(a) 2(x + 4) = 12 \n(b) 3(n – 5) = 21 \n(c) 3(n – 5) = – 21 \n(d) – 4(2 + x) = 8 \n(e) 4(2 – x) = 8 \nAnswer: \n(a) 2(x + 4) = 12 \nDividing both sides by 2, we get \n\\(\\frac{2(x+4)}{2}=\\frac{12}{2}\\) \u21d2 x + 4 = 6 \nTransposing 4 to R.H.S \nx = 6 – 4 = 2<\/p>\n
(b) 3(n – 5) = 21 \nDividing both sides by 3, we get \n\\(\\frac{3(n-5)}{3}=\\frac{21}{3}\\) \n\u21d2 n – 5 = 7 \nTransposing – 5 to R.H.S \nn = 7 + 5 = 12<\/p>\n
(c) 3(n – 5) = -21 \nDividing both sides by 3, we get \n\\(\\frac{3(n-5)}{3}=\\frac{-21}{3}\\) \n\u21d2 n – 5 = -7 \nTransposing – 5 to R.H.S \nn = -7 + 5 = -2<\/p>\n
(d) – 4(2 + x) =8 \nDividing both sides by -4, we get \n\\(\\frac{-4(2+x)}{-4}=\\frac{8}{-4}\\) \n\u21d2 2 + x = – 2 \nTransposing 2 to R.H.S \nx = -2 -2 = -4<\/p>\n
(e) 4(2 – x) = 8 \nDividing both sides by 4, we get 4(2-x) _ 8 \n\\(\\frac{4(2-\\mathrm{x})}{4}=\\frac{8}{4}\\) \n\u21d2 2 – x = 2 \nTransposing 2 to R.H.S \n-x = 2 – 2 = 0 \nMultiplying both sides by (- 1), we get \n-x \u00d7 (-1) = 0 \u00d7 (- 1) \nx = 0<\/p>\n
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Question 3. \nSolve the following equations: \n(a) 4 = 5(p – 2) \n(b) – 4 = 5(p – 2) \n(c) 16 = 4 + 3(t + 2) \n(d) 4 + 5(p – 1) = 34 \n(e) 0 = 16 + 4(m – 6) \nAnswer: \n(a) 4 = 5 (p – 2) \nInterchanging the sides, we get 5 ( p – 2) =4 \nDividing both sides by 5, we get \n\\(\\frac{5(p-2)}{5}=\\frac{4}{5}\\) \n\u21d2 p – 2 = \\(\\frac{4}{5}\\) \nTransposing -2 to R.H.S \np = \\(\\frac{4}{5}\\) + 2 \n= \\(\\frac{4+10}{5}=\\frac{14}{5}\\) = 2\\(\\frac{4}{5}\\)<\/p>\n
(b) – 4 = 5(p – 2) \nInterchanging the sides, we get } \n5 ( p – 2) = -4 \nDividing both sides by 5, we get \n\\(\\frac{5(p-2)}{5}=\\frac{-4}{5}\\) \np – 2 = \\(\\frac{-4}{5}\\) \nTransposing – 2 to R.H.S \np = \\(\\frac{-4}{5}\\) + 2 = \\(\\frac{-4+10}{5}\\) \n= \\(\\frac{6}{5}\\) = 1\\(\\frac{1}{5}\\)<\/p>\n
(c) 16 = 4 + 3(t + 2) \nInterchanging the sides, we get 4 + 3 (t + 2) = 16 \nTransposing 4 to R.H.S \n3(t + 2) – 16 – 4 = 12 \nDividing both sides by 3, we get \n\\(\\frac{3(t+2)}{3}=\\frac{12}{3}\\) \n\u21d2 t + 2 = 4 \nTransposing 2 to R.H.S \nt = 4 – 2 = 2<\/p>\n
(d) 4 + 5 (p – 1) = 34 \nTransposing 4 to R.H.S \n5 (p – 1) = 34 – 4 = 3 \nDividing both sides by 5, we get \n\\(\\frac{5(\\mathrm{p}-1)}{5}=\\frac{30}{5}\\) \np – 1 = 6 \nTransposing -1 to R.H.S \np = 6 + 1 = 7<\/p>\n
(e) 0 = 16 + 4(m – 6) \nInterchanging the sides we get \n16 + 4(m – 6) = 0 \nTransposing 16 to R.H.S \n4(m – 6) = 0 – 16 = – 16 \nDividing both sides by 4 we get \n\\(\\frac{4(\\mathrm{~m}-6)}{4}=\\frac{-16}{4}\\) \n\u21d2 m – 6 = – 4 \nTransposing – 6 to R.H.S \nm = -4 + 6 = 2<\/p>\n
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Question 4. \n(a) Construct 3 equations starting with x = 2 \n(b) Construct 3 equations starting with x = – 2 \nAnswer: \n(a) Starting with x = 2 \n(1) Multiplying both sides by 3, we get \n3x = 6 \nSubtracting 7 from both sides, we get \n3x – 7 = 6 – 7 \n3x – 7 = – 1<\/p>\n
(2) x = 2 \nAdding 5 on both sides, we get \nx + 5 = 2 + 5 \nx + 5 = 7 \nMultiplying both sides by 4, we get \n4(x + 5) = 7 \u00d7 4 \n4 (x + 5) = 28<\/p>\n
(b) Starting with x = – 2 \n(1) x = – 2 \nAdding 11 to both sides, we get \nx + 11 = – 2 + 11 \nx + 11 = 9<\/p>\n
(2) x = – 2 \nMultiplying both sides by 5, we get \n5x = – 10 \nAdding 3 on both sides, we get \n5x + 3 = – 10 + 3 \n5x + 3 = – 7<\/p>\n
(3) x = 2 \nDividing both sides by 5, we get \n\\(\\frac{x}{5}=\\frac{2}{5}\\) \nSubtracting 2 from both sides, we get \n\\(\\frac{\\mathrm{x}}{5}\\) – 2 = \\(\\frac{2}{5}\\) -2 \n\\(\\frac{\\mathrm{x}}{5}\\) – 2 = \\(\\frac{2-10}{5}\\) \n\\(\\frac{\\mathrm{x}}{5}\\) – 2 = \\(\\frac{-8}{5}\\)<\/p>\n
(3) x = – 2 \nDividing both sides by 7, we get \n\\(\\frac{x}{7}=\\frac{-2}{7}\\) \nAdd 3 on both sides, we get \n\\(\\frac{\\mathrm{X}}{7}\\) + 3 = \\(\\frac{-2}{7}\\) + 3 \n= \\(\\frac{-2+21}{7}\\) \n\\(\\frac{\\mathrm{X}}{7}\\) + 3 = \\(\\frac{19}{7}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3 Question 1. Solve the following equations: (a) 2y + (b) 5t + 28 = 10 (c) + 3 = …<\/p>\n
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3<\/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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 - 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