{"id":562,"date":"2020-09-15T09:46:16","date_gmt":"2020-09-15T04:16:16","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=562"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1<\/h2>\n

Question 1.
\nAftab tells his daughter, \u201cSeven years ago, I was, seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn\u2019t this interesting?) Represent this situation algebraically and graphically.
\nSolution:
\nLet the present age of father be x’
\nLet the age of daughter be ‘y’.
\nBefore seven years, age of father is x – 7
\nBefore seven years, age of daughter is y – 7.
\nFather was 7 times as old as daughter.
\n\u2234 x – 7 = 7 (y – 7)
\nx – 7 = 7y – 49
\nx – 7y = -49 + 7
\nx – 7y = – 42
\n\u2234 x – 7y + 42 = 0 …………….. (i)
\nAfter three years, age of father is x + 3
\nAfter three years, age of daughter is y + 3
\nThree years from now will three times as old as father.
\n\u2234 x + 3 = 3(y + 3)
\nx + 3 = 3y + 9
\nx – 3y + 3 – 9 = 0
\n3 – 3y – 6 = 0 ………….. (ii)
\n\u2234 Algebraically linear equations :
\nx – 7y + 42 = 0
\nx – 3y – 6 =0
\nIf we represent this linear equations through graph, we have
\ni) x – 7y + 42 = 0
\n-7y = -x – 42
\n7y = x + 42
\n\\(\\quad y=\\frac{x+42}{7}\\)<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n7<\/td>\n<\/tr>\n
\\(y=\\frac{x+42}{7}\\)<\/td>\n6<\/td>\n7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ii) x – 3y – 6 = 0
\n-3y = -x + 6
\n3y = x – 6
\n\\(\\quad y=\\frac{x-6}{3}\\)<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n+6<\/td>\n<\/tr>\n
\\(y=\\frac{x-6}{3}\\)<\/td>\n-2<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

Question 2.
\nThe coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebriacally and geometrically.
\nSolution:
\nLet the cost of each bat be Rs. ‘x’
\nLet the cost of each ball be Rs. ‘y’.
\n3x + 6y = 3900
\n\u2234 x + 2y = 1300 ………. (i)
\nCost of another bat and ball be x + 3y = 1300 ………….. (ii)
\n\u2234Algebriacally
\nx + 2y = 1300
\nx + 3y = 1300
\nAnd to represent through graph :
\ni) x + 2y = 1300
\n2y = 1300 – x
\n\\(\\quad x=\\frac{1300-x}{2}\\)<\/p>\n\n\n\n\n
x<\/td>\n100<\/td>\n200<\/td>\n<\/tr>\n
\\(y=\\frac{1300-x}{2}\\)<\/td>\n600<\/td>\n500<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ii) x + 3y = 1300
\n3y = 1300 – x
\n\\(\\quad y=\\frac{1300-x}{3}\\)<\/p>\n\n\n\n\n
x<\/td>\n100<\/td>\n400<\/td>\n<\/tr>\n
\\(y=\\frac{1300-x}{3}\\)<\/td>\n400<\/td>\n900<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

Question 3.
\nThe cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
\nSolution:
\nAlgebraically,
\nIf Cost of each kg. of apple be Rs. ‘x’,
\nCost of grapes be Rs. ‘y’. Then
\n2x + y = 160
\n4x + 2y = 300
\nTo represent geometrically,
\ni) 2x + y = 160
\ny = -2x + 160<\/p>\n\n\n\n\n
x<\/td>\n20<\/td>\n80<\/td>\n<\/tr>\n
y = 2x + 160<\/td>\n120<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ii) 4x + 2y = 300
\n2x + y= 150
\ny = 150 – 2x<\/p>\n\n\n\n\n
x<\/td>\n40<\/td>\n60<\/td>\n<\/tr>\n
y = 150 – 2x<\/td>\n70<\/td>\n30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB
\nWe hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1. Karnataka SSLC Class 10 Maths Solutions Chapter 3 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/562"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=562"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/562\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=562"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=562"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}