{"id":4747,"date":"2020-09-09T11:46:13","date_gmt":"2020-09-09T06:16:13","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4747"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-8-ex-8-1\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1"},"content":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1<\/h2>\n

1. Solve the following:<\/p>\n

Question i.
\nx + 3 = 11
\nAnswer:
\nx + 3 = 11
\nx = 11 – 3
\nx = 8<\/p>\n

Question ii.
\ny – 9 = 21
\nAnswer:
\ny – 9 = 21
\ny = 21 + 9
\ny = 30<\/p>\n

Question iii.
\n10 = z + 3
\n10 = z + 3 10-3 =z
\n7 = z or z = 7<\/p>\n

Question iv.
\n\\(\\frac{3}{11}+x=\\frac{9}{11}\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question v.
\n10x = 30
\nAnswer:
\n10x = 30
\nx = \\(\\frac { 30 }{ 10 }\\)
\nx = 3<\/p>\n

Question vi.
\n\\(\\frac{s}{7}=4\\)
\nAnswer:
\n\\(\\frac{s}{7}=4\\)
\nS = 4 \u00d7 7
\nS = 28<\/p>\n

Question vii.
\n\\(\\frac{3 x}{6}=10\\)
\nAnswer:
\n\\(\\frac{3 x}{6}=10\\)
\n3x = 10 \u00d7 6
\n3x = 60
\nx = \\(\\frac { 60 }{ 3 }\\)
\nx = 20<\/p>\n

Question viii.
\n\\(1.6=\\frac{x}{1.5}\\)
\nAnswer:
\n\\(1.6=\\frac{x}{1.5}\\)
\n1.6 \u00d7 1.5 = x
\n2.40 = x
\nx = 2.4<\/p>\n

Question ix.
\n8x – 8 = 48
\nAnswer:
\n8x – 8 = 48
\n8x = 48 + 8
\n8x = 56
\nx = \\(\\frac { 56 }{ 8 }\\)
\nx = 7<\/p>\n

Question x.
\n\\(\\frac{x}{3}+1=\\frac{7}{15}\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question xi.
\n\\(\\frac{x}{5}=12\\)
\nAnswer:
\n\\(\\frac{x}{5}=12\\)
\nx = 12 \u00d7 5
\nx = 60<\/p>\n

Question xii.
\n\\(\\frac{3 x}{5}=15\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question xiii.
\n3(x + 6) = 24
\nAnswer:
\n3(x + 6) = 24
\n3x+ 18 = 24
\n3x = 24 – 18
\n3x = 6
\nx = \\(\\frac { 6 }{ 3 }\\) = 2<\/p>\n

Question xiv.
\n\\(\\frac{x}{4}-8=1\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question xv.
\n3(x+2) – 2(x\u20141) = 7
\nAnswer:
\n3(x + 2) – 2(x – 1) = 7
\n3x + 6 – 2x + 2 = 7
\nx + 8 = 7 .
\nx = 7 – 8
\nx = -1<\/p>\n

2. Solve the equations :<\/p>\n

Question i.
\n5x = 3x + 24
\nAnswer:
\n5x = 3x + 24
\n2x = 24
\nx = \\(\\frac { 24 }{ 2 }\\)
\nx = 12<\/p>\n

Question ii.
\n8t + 5 = 2t – 31
\nAnswer:
\n8t + 5 = 2t – 31
\n6t = -36
\nt = \\(\\frac { -36 }{ 6 }\\)
\nt = -6<\/p>\n

Question iii.
\n7x -10 = 4x + 11
\nAnswer:
\n7x – 10 = 4x + 11
\n7x – 4x = 11 + 10
\n3x = 21
\nx = \\(\\frac { 21 }{ 3}\\)
\nx = 7<\/p>\n

Question iv.
\n4z + 3 = 6 + 2z
\nAnswer:
\n4z + 3 = 6 + 2z
\n4z – 2z = 6 – 3
\n2z = 3<\/p>\n

Question v.
\n2x – 1 = 14 – x
\nAnswer:
\n2x – 1 = 14 – x
\n2x + x = 14 + 1
\n3x = 15
\nx = \\(\\frac { 15 }{ 3 }\\)
\nx = 5<\/p>\n

Question vi.
\n6x + 1 = 3(x -1) + 7
\nAnswer:
\n6x + 1 = 3(x \u20141) + 7
\n6x + l =3x-3 + 7
\n6x + l = 3x + 4
\n6x – 3x = 4 – 1
\n3x = 3
\nx = \\(\\frac { 3 }{ 3 }\\)
\nx = 1<\/p>\n

Question vii.
\n\\(\\frac{2 x}{5}-\\frac{3}{2}=\\frac{x}{2}+1\\)
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question viii.
\n\\(\\frac{x-3}{5}-2=\\frac{2 x}{5}\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question ix.
\n3(x + 1) = 12 + 4(x – 1)
\nAnswer:
\n3(x + 1)= 12 + 4(x – 1)
\n3x + 3 = 12 + 4x – 4
\n3x + 3 = 4x + 8
\n3x – 4x = 8 – 3
\n-x = 5
\nx = -5<\/p>\n

Question x.
\n2x – 5 = 3(x – 5)
\nAnswer:
\n2x – 5 = 3(x – 5)
\n2x – 5 = 3x – 15
\n2x – 3x = -15 + 5
\n-x = -10
\nx= 10<\/p>\n

Question xi.
\n6(1 – 4x) + 7(2 + 5x) = 53
\nAnswer:
\n6( 1 – 4x) + 7(2 + 5x) = 53
\n6 – 24x+ 14 + 35x = 53
\n35x – 24x + 6 + 14 = 53
\n11x+ 20 = 53
\n11x = 53 – 20
\nx = \\(\\frac { 33 }{ 11 }\\)
\nx = 3<\/p>\n

Question xii.
\n3(x + 6) + 2(x + 3) = 64
\nAnswer:
\n3(x + 6) + 2(x + 3) = 64
\n3x + 18 + 2x + 6 = 64
\n3x + 2x + 18 + 6 = 64
\n5x + 24 = 64
\n5x = 64 – 24
\n5x = 40
\nx = \\(\\frac { 40 }{ 5 }\\)
\n\u2234 x = 8<\/p>\n

Question xiii.
\n\\(\\frac{2 m}{3}+8=\\frac{m}{2}-1\\)
\nAnswer:
\n\"KSEEB<\/p>\n

Question xiv.
\n\\(\\frac{3}{4}(x-1)=(x-3)\\)
\nAnswer:
\n\\(\\frac{3}{4}(x-1)=(x-3)\\)
\n3(x – 1) = 4(x – 3)
\n3x – 3 = 4x – 12
\n3x – 4x = -12 + 3
\n-x = -9
\nx = 9<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4747"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=4747"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4747\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4747"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4747"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4747"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}