{"id":693,"date":"2020-09-15T05:18:47","date_gmt":"2020-09-14T23:48:47","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=693"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-4\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4<\/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 3.4.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4<\/h2>\n

Question 1.
\nSolve the following pair of linear equations by the elimination method and the substitution method :
\n(i) x + y = 5 and 2x – 3y = 4
\n(ii) 3x + 4y =10 and 2x – 2y = 2
\n(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
\n(iv) \\(\\frac{x}{2}+\\frac{2 y}{3}=-1 \\quad \\text { and } \\quad x-\\frac{y}{3}=3\\)
\nSolution:
\n(i) x + y = 5 and 2x – 3y = 4
\nx + y = 5 ………….. (i)
\n2x – 3y = 4 ………… (ii)
\nMultiplying eqn. (i) by 3
\n3x + 3y = 15 ………….. (iii)
\nBy adding eqn. (ii) to eqn. (iii) ‘y’ is eliminated.
\n\"KSEEB
\n\u2234 \\( x=\\frac{19}{5}\\)
\nSubstituting the value of ‘x’ in eqn. (i),
\nx + y = 5
\n\\(\\frac{19}{5}+y=5\\)
\n\\(y=\\frac{5}{1}-\\frac{19}{5}\\)
\n\\(=\\frac{25-19}{5}\\)
\n\\( \\quad y=\\frac{6}{5}\\)
\n\\( \\quad x=\\frac{19}{5} y=\\frac{6}{5}\\)<\/p>\n

(ii) 3x + 4y = 10 and 2x – 2y = 2
\n3x + 4y = 10 ……….. (i)
\n2x – 2y = 2 …………. (ii)
\nMultiplying eqn. (ii) by 2,
\n4x – 4y = 4 ………….. (iii)
\nAdding eqn. (i) to eqn. (ii),
\n\"KSEEB
\n\u2234 \\(\\quad x=\\frac{14}{7}=2\\)
\nSubstituting the value of ‘x’ in eqn. (i),
\n3x + 4y = 10
\n3 \u00d7 2 + 4y = 10
\n6 + 4y = 10
\n4y = 10 – 6
\n4y = 4
\n\u2234 \\(\\mathrm{y}=\\frac{4}{4}=1\\)
\n\u2234 x = 2, y = 1.<\/p>\n

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
\n3x – 5y – 4 = 0 \u21d2 3x – 5y = 4 ……… (i)
\n9x = 2y + 7 \u21d2 9x – 2y = 7 ………. (ii)
\nMultiplying eqn. (i) by 3,
\n9x – 15y = 12 …………. (iii)
\nSubtracting eqn. (iii) from eqn. (ii)
\n\"KSEEB
\n\u2234 \\(\\quad y=\\frac{-5}{13}\\)
\nSubstituting the value of ‘y’ in eqn. (i),
\n3x – 5y = 4
\n\\(3 x-5\\left(\\frac{-5}{13}\\right)=4\\)
\n\\(3 x+\\frac{25}{13}=4 \\quad \\text { OR } \\quad 3 x=4-\\frac{25}{13}\\)
\n\\(3 x=\\frac{52-25}{13}=\\frac{27}{13}\\)
\n\\(3 x=\\frac{27}{13}\\)
\n\u2234 \\(\\quad x=\\frac{27}{13} \\times \\frac{1}{3}\\)
\n\u2234 \\(\\quad x=\\frac{9}{13}\\)
\n\u2234 \\(\\quad x=\\frac{9}{13} \\quad y=\\frac{-5}{13}\\)<\/p>\n

(iv) \\(\\frac{x}{2}+\\frac{2 y}{3}=-1 \\text { and } x-\\frac{y}{3}=3\\)
\n\\(\\frac{x}{2}+\\frac{2 y}{3}=-1 \\quad \\frac{x}{1}-\\frac{y}{3}=3\\)
\n\\(\\frac{3 x+4 y}{6}=-1 \\quad \\frac{3 x-y}{3}=3\\)
\n3x + 4y = -6 ……….. (i) 3x – y = 9 …………… (ii)
\nSubtracting eqn. (ii) from eqn. (i),
\n\"KSEEB
\n\\(y=\\frac{-15}{3}=-3\\)
\ny = -3
\nSubstituting the value of ‘y’ in eqn. (i),
\n3x + 4y = -6
\n3x + 4 (-3) = -6
\n3x – 12 = -6
\n3x = -6 + 12
\n3x = 6
\n\u2234 \\(\\quad x=\\frac{6}{3} \\quad=2\\)
\n\u2234 x = 2, y = 3<\/p>\n

Question 2.
\nForm the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:<\/p>\n

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \\(\\frac{1}{2}\\) if we only add 1 to the denominator. What is the fraction ?
\nSolution:
\nLet the fraction be \\(\\frac{x}{y}\\), Here numerator is x and denominator is y.
\n\u2234 \\(\\quad \\frac{\\mathrm{x}+1}{\\mathrm{y}-1}=1\\)
\nx + 1 = y – 1
\nx – y = -1 – 1
\nx – y = -2 ……….. (i)
\n\\(\\frac{x}{y+1}=\\frac{1}{2}\\)
\n2 \u00d7 x = y + 1
\n2x = y + 1
\n2x – y = 1 ………. (ii)
\nSubtracting eqn. (ii) from eqn. (i),
\n\"KSEEB
\n\u2234 x = 3
\nSubstituting the value of ‘x’ in eqn. (i),
\nx – y = -2
\n3 – y = -2
\n-y = -2 -3
\n-y = -5
\n\u2234 y = 5
\n\u2234 Fraction is \\(\\frac{x}{y}=\\frac{3}{5}\\)<\/p>\n

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
\nSolution:
\nPresent age of Nuri is ‘x’
\nPresent age of Sonu is ‘y’
\nFive years ago, age of Nuri is x – 5
\nFive years ago, age of Sonu is y – 5
\nthen, x – 5 = 3(y – 5)
\nx – 5 = 3y – 15
\nx – 3y = -15 + 5
\nx – 3y = -10 ……….. (i)
\nAfter 10 years, age of Nuri will be x + 10
\nAfter 10 years, age of Sonu will be y + 10
\nthen, x +10 = 2(y + 10)
\nx + 10 = 2y + 20
\nx – 2y = 20 – 10
\nx – 2y = 10 …………. (ii)
\nSubtracting eqn. (ii) in eqn. (i),
\n\"KSEEB
\n\u2234 y = 20
\nSubstituting the value of ‘y’ in eqn. (ii),
\nx – 2y = 10
\nx – 2(20) = 10
\nx – 40 = 10
\nx = 10 + 40
\nx = 50
\n\u2234 Present age of Nuri, x = 50
\nPresent age of Sonu, y = 20.<\/p>\n

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
\nSolution:
\nLet the two digit number is 10x + y.
\nSum of the digits of to digit number is
\nx + y = 9 ………….. (i)
\nNumber obtained by reversing the order of the digits, we get 10y + x.
\nTwicing the 10y + x, it is nine times of first number.
\n\u2234 2(10y + x) = 9(10x + 4)
\n20y + 2x = 90x + 9y
\n2x – 90x + 20y – 9y = 0
\n-88x + 11y = 0
\n88x – 11y = 0
\n8x – y = 0 ………….. (ii)
\nBy adding eqn. (i) to eqn. (ii),
\n\"KSEEB
\n\u2234 \\(x=\\frac{9}{9}=1\\)
\nSubstituting the value of ‘x’ in eqn. (i),
\nx + y = 9
\n1 + y = 9
\ny = 9 – 1
\ny = 8
\n\u2234 x = 1, y = 8
\n\u2234 Two digit number = 10x + y
\n= 10 \u00d7 1 + 8
\n= 10 + 8
\n=18.<\/p>\n

(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
\nSolution:
\nLet the Rs. 50 notes be ‘x’
\nRs. 100 notes b ‘y’
\nx + y = 25 …………. (i)
\nx + 2y = 40 …………. (ii)
\nSubtracting eqn. (ii) from eqn. (i),
\n\"KSEEB
\n\u2234 y = 15
\nSubstituting the value of ‘y’ in eqn. (i)
\nx + y = 25
\nx+ 15 = 25
\nx = 25 – 15
\n\u2234 x = 10
\n\u2234 x = 10, y = 15
\n\u2234 Rs. 50’s notes are 10 and
\nRs. 100’s notes are 15.<\/p>\n

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
\nSolution:
\nFixed charge be Rs. x.
\nAdditional charge be Rs. y.
\nx + 4y = 27 ………. (i)
\nx + 2y = 21 ………. (ii)
\nSubtracting Eqn. (ii) from eqn. (i),
\n\"KSEEB
\n\u2234 \\(\\quad y=\\frac{6}{2} \\quad=3\\)
\nSubstituting the value of ‘y’ in eqn. (i)
\nx + 4y = 27
\nx + 4(3) = 27
\nx + 12 = 27
\n\u2234 x = 27 – 12
\n\u2234 x = 15
\n\u2234 Fixed charge, x = Rs. 15.
\nAdditional charge, y = Rs. 3.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 will help you. If you have any query regarding Karnataka Board SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4, 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 3.4 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 3.4. 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\/693"}],"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=693"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/693\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=693"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=693"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=693"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}