Solve the following equations.<\/span><\/p>\nQuestion 1. \n\\(\\frac{8 x-3}{3 x}=2\\) \nSolution: \n\\(\\frac{8 x-3}{3 x}=2\\) \nBy cross-multiplication we get, \n8x – 3 = 2(3x) \n8x – 3 = 6x \nBy transposing -3 to the R.H.S. and 6x to L.H.S. \n8x – 6x = 3 \n2x = 3 \nDividing both sides by 2 \n\\(\\frac{2 x}{2}=\\frac{3}{2}\\) \n\u2234 x = \\(\\frac{3}{2}\\)<\/p>\n
Question 2. \n\\(\\frac{9 x}{7-6 x}=15\\) \nSolution: \n\\(\\frac{9 x}{7-6 x}=15\\) \nBy cross-multiplication, we get \n9x = 15(7 – 6x) \n9x = 105 – 90x \nBy transposing -90x to L.H.S., we get \n9x + 90x = 105 \n99x = 105 \nDividing both sides by 99, we get \n\\(\\frac{99 \\mathrm{x}}{99}=\\frac{105}{99}\\) \n\u2234 x = \\(\\frac{35}{33}\\)<\/p>\n
Question 3. \n\\(\\frac{z}{z+15}=\\frac{4}{9}\\) \nSolution: \n\\(\\frac{z}{z+15}=\\frac{4}{9}\\) \nBy cross-multiplication, we get \n9z = 4(z + 15) \n9z = 4z + 60 \nTransposing 4z to L.H.S, we get \n9z – 4z = 60 \n5z = 60 \nDividing both sides by 5, we get \n\\(\\frac{5 z}{5}=\\frac{60}{5}\\) \n\u2234 z = 12<\/p>\n
Question 4. \n\\(\\frac{3 y+4}{2-6 y}=\\frac{-2}{5}\\) \nSolution: \n\\(\\frac{3 y+4}{2-6 y}=\\frac{-2}{5}\\) \nBy cross-multiplication, we get \n5(3y + 4) = -2(2 – 6y) \n15y + 20 = -4 + 12y \nTransposing 12y to L.H.S. and 20 to R.H.S. \n15y – 12y = -4 – 20 \n3y = -24 \nDividing both sides by 3, we get \n\\(\\frac{3 y}{3}=-\\frac{24}{3}\\) \n\u2234 y = -8<\/p>\n
Question 5. \n\\(\\frac{7 y+4}{y+2}=\\frac{-4}{3}\\) \nSolution: \n\\(\\frac{7 y+4}{y+2}=\\frac{-4}{3}\\) \nBy cross-multiplication, we get \n3(7y + 4) = -4(y + 2) \n21y + 12 = -4y – 8 \nTransposing +12 to R.H.S. and -4y to L.H.S. \n21y + 4y = -8 – 12 \n25y = -20 \nDividing both sides by 25, we get \n\\(\\frac{25 y}{25}=\\frac{-20}{25}\\) \n\u2234 y = \\(-\\frac{4}{5}\\)<\/p>\n
Question 6. \nThe ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages. \nSolution: \nLet the present age of Hari = 5x years \nand the present age of Harry = 7x years \nAfter four years \nAge of Hari = (5x + 4) years \nAge of Harry = (7x + 4) years \nAccording to the question \n(5x + 4) : (7x + 4) = 3 : 4 \n4(5x + 4) = 3(7x + 4) \n[Product of the extremes is equal to the product of the means] \n20x + 16 = 21x + 12 \nTransposing 16 to R.H.S. and 21x to L.H.S., we get \n20x – 21x = 12 – 16 \n-x = -4 \n\u2234 x = 4 \nPresent age of Hari = 5 \u00d7 4 = 20 years \nPresent age of Harry = 7 \u00d7 4 = 28 years<\/p>\n
Question 7. \nThe denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \\(\\frac{3}{2}\\). Find the rational number. \nSolution: \nLet the numerator be ‘x’ \nthen the denominator is x + 8 \n\u2234 The fraction is \\(\\frac{\\mathrm{x}}{\\mathrm{x}+8}\\) \nAccording to the question, we get \n\\(\\frac{x+17}{(x+8)-1}=\\frac{3}{2}\\) \n\\(\\frac{x+17}{x+8-1}=\\frac{3}{2}\\) \n\\(\\frac{x+17}{x+7}=\\frac{3}{2}\\) \nBy cross-multiplication, we get \n3(x + 7) = 2(x + 17) \n3x + 21 = 2x + 34 \nBy transposing 21 to R.H.S. and 2x to L.H.S., we get \n3x – 2x = 34 – 21 \nx = 13 \n\u2234 The fraction is \\(\\frac{13}{13+8}\\) = \\(\\frac{13}{21}\\) \nor The rational number = \\(\\frac{13}{21}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 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 2.6 Solve the following equations. Question 1. Solution: By cross-multiplication we get, 8x – …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6<\/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 2.6 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n