{"id":27128,"date":"2021-06-29T15:22:28","date_gmt":"2021-06-29T09:52:28","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27128"},"modified":"2021-06-29T16:01:27","modified_gmt":"2021-06-29T10:31:27","slug":"ncert-solutions-for-class-7-maths-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-4-ex-4-3\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3"},"content":{"rendered":"

These 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>\n

NCERT 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

\"NCERT<\/p>\n

(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

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

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

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3 Question 1. 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