{"id":2164,"date":"2020-09-08T10:19:48","date_gmt":"2020-09-08T04:49:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2164"},"modified":"2021-07-02T15:03:43","modified_gmt":"2021-07-02T09:33:43","slug":"kseeb-solutions-class-9-maths-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-10-ex-10-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2<\/strong>\u00a0are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.2.<\/p>\n

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2<\/h2>\n

Question 1.
\nWhich one of the following options is true, and why ?
\ny = 3x + 5 has
\ni) a unique solution,
\nii) only two solutions,
\niii) infinitely many solutions.
\nSolution:
\ny = 3x + 5 has many solutions. Because this is a linear equation with two variables.<\/p>\n

Question 2.
\nWrite four solutions for each of the following equations :
\ni) 2x + y = 7
\nii) \u03c0x + y = 9
\niii) x = 4y
\nSolution:
\ni) 2x + y = 7
\nIf x = 0, 2x + y = 7
\n2 \u00d7 0 + y = 7
\n0 + y = 7
\ny = 7
\nSolution is (0, 7).
\nIf x = 1, 2x + y = 7
\n2 \u00d7 1 + y = 7
\n2 + y = 7
\ny = 7 – 2 = 5
\nSolution is (1, 5).
\nIf x = 2, 2x + y = 7
\n2 \u00d7 2 + y = 7
\n4 + y = 7
\ny = 7 – 4 = 3
\nSolution is (2, 3).
\nIf x = 3, 2x + y = 7
\n2 \u00d7 3 + y = 7
\n6 + y = 7
\n\u2234 y = 7 – 6 = 1
\nSolution is (3, 1).<\/p>\n

ii) \u03c0x + y = 9
\nIf x = 0, \u03c0x + y = 9
\n\u03c0 \u00d7 0 + y = 9
\n0 + y = 9
\ny = 9
\nSolution (x, y) = (0, 0).
\nIf x = 1, \u03c0x + y = 9
\n\u03c0 \u00d7 1 + y = 9
\n\u03c0 + y = 9
\ny = 9 – \u03c0
\nSolution (x, y) = (1, 9 – \u03c0).
\nIf y = 0, \u03c0x + y = 9
\n\u03c0x + 0 = 9
\n\u03c0x = 9
\n\u2234 x = \\(\\frac{9}{\\pi}\\)
\nSolution (x, y) = ( \\(\\frac{9}{\\pi}\\), o).
\nIf y = 1, \u03c0x + y = 9
\n\u03c0x + 1 = 9
\n\u03c0x = 9 – 1
\n\u03c0x = 8
\n\u2234 x = \\(\\frac{8}{\\pi}\\)
\nSolution (x, y) = (\\(\\frac{8}{\\pi}\\) ,1).<\/p>\n

iii) x = 4y
\nIf x = 0, x = 4y
\n0 = 4y
\n4y = 0
\n\u2234 y = \\(\\frac{0}{4}\\) = \u221e
\n\u2234 Solution (x, y) = (0, \u221e).
\nIf x = 1, x = 4y
\n1 = 4y
\n4y = 1
\n\u2234 y = \\(\\frac{1}{4}\\)
\n\u2234 Solution (x, y) = (1, \\(\\frac{1}{4}\\)).
\nIf x = 2, x = 4y
\n2 = 4y
\n4y = 2
\n\u2234 y = \\(\\frac{4}{2}\\)
\n\u2234 y = \\(\\frac{1}{2}\\)
\nSolution (x, y) = (2, \\(\\frac{1}{2}\\)).
\nIf x = 4, x = 4y
\n4 = 4y
\n4y = 4
\n\u2234 y = \\(\\frac{4}{4}\\)
\n\u2234 y = 1
\n\u2234 Solution (x, y) = (4, 1).<\/p>\n

Question 3.
\nCheck which of the following are solutions of the equation x – 2y = 4 and which are not:
\ni) (0, 2)
\nii) (2, 0)
\niii) (4, 0)
\niv) (\\(\\sqrt{2}, 4 \\sqrt{2}\\))
\nv) (1, 1)
\nSolution:
\ni) (0, 2)
\nx – 2y = 4
\n0 – 2(2) = 4
\n0 – 4 = 4
\n-4 \u2260 4
\n\u2234 (0, 2) is not a solution.<\/p>\n

ii) (2, 0)
\nx – 2y = 4
\n2 – 2(0) = 4
\n2 – 0 = 4
\n2 \u2260 4
\n\u2234 (2, 0) is not a solution<\/p>\n

iii) (4, 0)
\nx – 2y = 4
\n4 – 2(0) = 4
\n4 – 0 = 4
\n4 = 4
\n\u2234 (4, 0) is a solution.<\/p>\n

iv)
\n\"KSEEB<\/p>\n

v) (1, 1)
\nx – 2y = 4
\n1 – 2(1) = 4
\n1 – 2 = 4
\n-1 \u2260 4
\n\u2234 (1, 1) is not a solution.<\/p>\n

Question 4.
\nFind the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
\nSolution:
\n2x + 3y = k
\nIf x = 2, y = 1, then k = ?
\n2x + 3y = k
\n2(2) + 3(1) = k
\n4 + 3 = k
\n7 = k
\n\u2234 k = 7.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter Linear Equations in Two Variables Ex 10.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter Linear Equations in Two Variables Exercise 10.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2\u00a0are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.2. Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2 …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2164"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2164"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2164\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2164"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2164"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}