Solve the following equations and check your results.<\/span><\/p>\nQuestion 1. \n3x = 2x + 18 \nSolution: \n3x = 2x + 18 \nTransposing 2x from RHS to L.H.S, we get \n3x – 2x = 18 \nx = 18<\/p>\n
Check: \nPut x = 18 in L.H.S. and R.H.S. of the equation. \nL.H.S. = 3x = 3 \u00d7 18 = 54 \nR.H.S. = 2x + 18 = 2(18) + 18 = 36 +18 = 54 \n\u2234 L.H.S. = R.H.S<\/p>\n
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Question 2. \n5t – 3 = 3t – 5 \nSolution: \n5t – 3 = 3t – 5 \nTransposing (-3) to R.H.S. and 3t to L.H.S., we get \n5t – 3t = -5 + 3 \n2t = -2 \nDividing both sides by 2, we get \n\\(\\frac{2 t}{2}=\\frac{-2}{2}\\) \nt = -1<\/p>\n
Check: \nPut t = -1 in L.H.S. and RHS of the equation \nL.H.S. = 5t – 3 = 5(-1) – 3 = -5 – 3 = -8 \nR.H.S. = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8 \nHence, L.H.S. = R.H.S.<\/p>\n
Question 3. \n5x + 9 = 5 + 3x \nSolution: \n5x + 9 = 5 + 3x \nTransposing 9 to R.H.S. and 3x to L.H.S. \n5x – 3x = 5 – 9 \n2x = -4 \nDividing both sides by 2, we get \n\\(\\frac{2 x}{2}=\\frac{-4}{2}\\) \nx = -2<\/p>\n
Check: \nPut x = -2 in L.H.S. and R.H.S. of the equation \nL.H.S. = 5x + 9 \n= 5 (-2) + 9 \n= -10 + 9 \n= -1 \nR.H.S. = 5 + 3x \n= 5 + 3(-2) \n= 5 – 6 = -1 \nHence, L.H.S. = R.H.S.<\/p>\n
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Question 4. \n4z + 3 = 6 + 2z \nSolution: \n4z + 3 = 6 + 2z \nTransposing 3 to R.H.S. and 2z to LHS \n4z – 2z = 6 – 3 \n2z = 3 \nDividing both sides by 2, we get \n\\(\\frac{2 z}{2}=\\frac{3}{2}\\) \nz = \\(\\frac{3}{2}\\)<\/p>\n
Check: \nPut z = \\(\\frac{3}{2}\\) in L.H.S. and R.H.S of the equation \nL.H.S. = 4z + 3 \n= 4\\(\\left(\\frac{3}{2}\\right)\\) + 3 \n= 2(3) + 3 \n= 6 + 3 \n= 9 \nR.H.S = 6 + 2z \n= 6 + 2\\(\\left(\\frac{3}{2}\\right)\\) \n= 6 + 3 \n= 9 \nHence, L.H.S. = R.H.S.<\/p>\n
Question 5. \n2x – 1 = 14 – x \nSolution: \n2x – 1 = 14 – x \nTransposing -1 to RHS and -x to L.H.S. \n2x + x = 14 + 1 \n3x = 15 \nDividing both sides by 3, we get \n\\(\\frac{3 x}{3}=\\frac{15}{3}\\) \nx = 5<\/p>\n
Check: \nPut x = 5 in LHS and RHS of the equation \nLHS = 2x – 1 \n= 2(5) – 1 \n= 10 – 1 \n= 9 \nR.H.S = 14 – x \n= 14 – 5 \n= 9 \nHence, L.H.S. = R.H.S.<\/p>\n
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Question 6. \n8x + 4 = 3(x – 1) + 7 \nSolution: \n8x + 4 = 3(x – 1) + 7 \n8x + 4 = 3x – 3 + 7 \n8x + 4 = 3x + 4 \nTransposing 4 to R.H.S. and 3x to LHS, we get \n8x – 3x = 4 – 4 \n5x = 0 \nDividing both sides by 5 \n\\(\\frac{5 x}{5}=\\frac{0}{5}\\) \nx = 0<\/p>\n
Check: \nPut x = 0 in L.H.S. and R.H.S. of the equation \nL.H.S. = 8x + 4 \n= 8(0) + 4 \n= 4 \nR.H.S = 3(x – 1) + 7 \n= 3(0 – 1) + 7 \n= -3 + 7 \n= 4 \nHence, L.H.S. = R.H.S.<\/p>\n
Question 7. \nx = \\(\\frac{4}{5}\\) (x + 10) \nSolution: \nx = \\(\\frac{4}{5}\\) (x + 10) \nMultiplying both sides by 5, we get \n5x = 5 \u00d7 \\(\\frac{4}{5}\\) (x + 10) \n5x = 4(x + 10) \n5x = 4x + 40 \nTransposing 4x to L.H.S. \n5x – 4x = 40 \nx = 40<\/p>\n
Check: \nPut x = 40 in L.H.S. and R.H.S. of the equation \nL.H.S. = x = 40 \nR.H.S. = \\(\\frac{4}{5}\\) (x + 10) \n= \\(\\frac{4}{5}\\) (40 + 10) \n= \\(\\frac{4}{5}\\) x 50 \n= 4 \u00d7 10 \n= 40 \nHence, L.H.S. = R.H.S<\/p>\n
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Question 8. \n\\(\\frac{2 x}{3}+1=\\frac{7 x}{15}+3\\) \nSolution: \n\\(\\frac{2 x}{3}+1=\\frac{7 x}{15}+3\\) \nTransposing 1 to R.H.S. and \\(\\frac{7 \\mathrm{x}}{15}\\) to L.H.S. \n\\(\\frac{2 x}{3}-\\frac{7 x}{15}=3-1\\) \n\\(\\frac{10 x-7 x}{15}\\) = 2 \n\\(\\frac{3 x}{15}\\) = 2 \nMultiplying both sides by 15, we get \n\\(\\frac{3 x}{15}\\) \u00d7 15 = 2 \u00d7 15 \n3x = 30 \nDividing both sides by 3 \n\\(\\frac{3 x}{3}=\\frac{30}{3}\\) \nx = 10<\/p>\n
Check: \nPut x = 10 in L.H.S. and R.H.S. of the equation \n \nHence, L.H.S. = R.H.S.<\/p>\n
Question 9. \n\\(2 y+\\frac{5}{3}=\\frac{26}{3}-y\\) \nSolution: \n\\(2 y+\\frac{5}{3}=\\frac{26}{3}-y\\) \nTransposing \\(\\frac{5}{3}\\) to R.H.S. and -y to L.H.S. \n2y + y = \\(\\frac{26}{3}-\\frac{5}{3}\\) \n3y = \\(\\frac{26-5}{3}\\) \n3y = \\(\\frac{21}{3}\\) = 7 \nDivide both sides by 3, we get \n\\(\\frac{3 y}{3}=\\frac{7}{3}\\) \ny = \\(\\frac{7}{3}\\)<\/p>\n
Check: \nPut y = \\(\\frac{7}{3}\\) in .L.H.S. and R.H.S of the equation \n \nHence, L.H.S. = R.H.S<\/p>\n
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Question 10. \n3m = 5m – \\(\\frac{8}{5}\\) \nSolution: \n3m = 5m – \\(\\frac{8}{5}\\) \nTransposing 5m to L.H.S. we get \n3m – 5m = \\(\\frac{-8}{5}\\) \n-2m = \\(\\frac{-8}{5}\\) \nDividing both sides by -2, we get \n\\(\\frac{-2 m}{-2}=\\frac{-8}{5} \\div(-2)\\) \nm = \\(\\frac{8}{10}=\\frac{4}{5}\\)<\/p>\n
Check: \nPut m = \\(\\frac{4}{5}\\) in L.H.S. and R.H.S. of the equation \n \nHence, L.H.S = R.H.S<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 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 2.3 Solve the following equations and check your results. Question 1. 3x = 2x …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3<\/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":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 - 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