{"id":26908,"date":"2021-06-28T16:29:15","date_gmt":"2021-06-28T10:59:15","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26908"},"modified":"2021-06-28T16:29:15","modified_gmt":"2021-06-28T10:59:15","slug":"ncert-solutions-for-class-8-maths-chapter-2-ex-2-6","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-6\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 2 Linear Equations in One Variable Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6<\/h2>\n

Solve the following equations.<\/span><\/p>\n

Question 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":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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