{"id":750,"date":"2020-09-15T09:02:24","date_gmt":"2020-09-15T03:32:24","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=750"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-6\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6<\/strong>\u00a0are 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 3.6.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6<\/h2>\n

Question 1.
\nSolve the following pairs of equations by reducing them to a pair of linear equations :
\n(i) \\(\\frac{1}{2 x}+\\frac{1}{3 y}=2\\)
\n\\(\\frac{1}{3 x}+\\frac{1}{2 y}=\\frac{13}{6}\\)
\n(ii) \\(\\frac{2}{\\sqrt{x}}+\\frac{3}{\\sqrt{y}}=2\\)
\n\\(\\frac{4}{\\sqrt{x}}+\\frac{9}{\\sqrt{y}}=-1\\)
\n(iii) \\(\\frac{4}{x}+3 y=14\\)
\n\\(\\frac{3}{x}-4 y=23\\)
\n(iv) \\(\\frac{5}{x-1}+\\frac{1}{y-2}=2\\)
\n\\(\\frac{6}{x-1}-\\frac{3}{y-2}=1\\)
\n(v) \\(\\frac{7 x-2 y}{x y}=5\\)
\n\\(\\frac{8 x+7 y}{x y}=15\\)
\n(vi) 6x + 3y = 6xy
\n2x + 4y = 5xy
\n(vii) \\(\\frac{10}{x+y}+\\frac{2}{x-y}=4\\)
\n\\(\\frac{15}{x+y}+\\frac{5}{x-y}=-2\\)
\n(viii) \\(\\frac{1}{3 x+y}+\\frac{1}{3 x-y}=\\frac{3}{4}\\)
\n\\(\\frac{1}{2(3 x+y)}-\\frac{1}{2(3 x-y)}=\\frac{-1}{8}\\)
\nSolution:
\n(i) \\(\\frac{1}{2 x}+\\frac{1}{3 y}=2\\) ……………. (i)
\n\\(\\frac{1}{3 x}+\\frac{1}{2 y}=\\frac{13}{6}\\) ……………… (ii)
\nFrom eqn. (i),
\n\\(\\frac{1}{2 x}+\\frac{1}{3 y}=2\\)
\n\\(\\frac{3 y+2 x}{6 x y}=2\\)
\n3y + 2x = 12xy
\n2x + 3y – 12xy = 0
\n2x + 3y = 12xy …………….. (i)
\nFrom eqn. (ii),
\n\\(\\frac{1}{3 x}+\\frac{1}{2 y}=\\frac{13}{6}\\)
\n\\(\\frac{2 y+3 x}{6 x y}=\\frac{13}{6}\\)
\n6(2y + 3x) = 13 \u00d7 6xy
\n12y + 18x = 78xy
\n18x + 12y – 78xy = 0
\n18x + 12y = 78xy ……………… (ii)
\nMultiplying eqn. (i) by 4,
\n4(2x + 3y) = 4 \u00d7 12xy
\n8x + 12y = 78xy …………….. (iii)
\nFrom eqn. (ii) – eqn. (iii),
\n\"KSEEB
\n\\(x=\\frac{30 x y}{10}=3 x y\\)
\n\\(\\frac{x}{x}=3 y\\)
\n1 = 3y
\n\\(y=\\frac{1}{3}\\)
\n2x + 3y = 12xy
\n\\(2 x+3\\left(\\frac{1}{3}\\right)=12 \\times x \\times \\frac{1}{3}\\)
\n2x + 1 = 4x
\n2x – 4x = 1
\n-2x = -1
\n2x = 1
\n\\(x=\\frac{1}{2}\\)
\n\\(x=\\frac{1}{2}, \\quad y=\\frac{1}{3}\\)<\/p>\n

(ii) \\(\\frac{2}{\\sqrt{x}}+\\frac{3}{\\sqrt{y}}=2\\) ………….. (i)
\n\\(\\frac{4}{\\sqrt{x}}+\\frac{9}{\\sqrt{y}}=-1\\) ………….. (ii)
\nFrom eqn (i)
\n\\(\\frac{2}{\\sqrt{x}}+\\frac{3}{\\sqrt{y}}=2\\)
\n\\(\\frac{2 \\sqrt{y}+3 \\sqrt{x}}{\\sqrt{x y}}=2\\)
\n\\(3 \\sqrt{x}+2 \\sqrt{x y}=2 \\sqrt{x y}\\) …………… (i)
\nFrom eqn (ii),
\n\\(\\frac{4}{\\sqrt{x}}+\\frac{9}{\\sqrt{y}}=-1\\)
\n\\(\\frac{4 \\sqrt{y}-9 \\sqrt{x}}{\\sqrt{x y}}=-1\\)
\n\\(-9 \\sqrt{x}+4 \\sqrt{y}=-\\sqrt{x y}\\)
\n\\(9 \\sqrt{x}-3 \\sqrt{y}=\\sqrt{x y}\\) ……………….. (ii)
\nMultiplying eqn (i) with 2
\n\\(2(3 \\sqrt{x}+2 \\sqrt{y})=2 \\times 2 \\sqrt{x y}\\)
\n\\(6 \\sqrt{x}+4 \\sqrt{y}=4 \\sqrt{x y}\\) ……………….. (iii)
\nBy adding eqn (ii) + eqn (iii)
\n\"KSEEB
\n\\(\\sqrt{x}=\\frac{5 \\sqrt{x y}}{15}\\)
\n\\(\\frac{\\sqrt{x}}{\\sqrt{x}}=\\frac{5 \\sqrt{y}}{15}\\)
\n\\(1=\\frac{\\sqrt{y}}{3}\\)
\n\\(3=\\sqrt{y} \\quad \\sqrt{y}=3\\)
\n\u2234 y = 9
\nSubstituting the value of ‘y’ in eqn (i)
\n\"KSEEB
\n\u2234 x = 4, y = 9<\/p>\n

(iii) \\(\\frac{4}{x}+3 y=14\\)
\n\\(\\frac{3}{x}-4 y=23\\)
\nby putting the value of By multiplying eqn. (i) by 4, multiplying eqn. (ii) by 3, and then adding them,
\n\"KSEEB
\n\u2234 p = 5
\n\\(\\frac{1}{x}=p=5 \\quad \\quad x=\\frac{1}{5}\\)
\nSubstituting the value of ‘p’ in eqn. (i)
\n4p + 3y = 14
\n4 \u00d7 5 + 3y = 14
\n20 + 3y = 14
\n3y = 14 – 20
\n3y = -6
\n\u2234 y = -2
\n\u2234 \\(\\quad x \\cdot=\\frac{1}{5}, y=-2\\)<\/p>\n

(iv) \\(\\frac{5}{x-1}+\\frac{1}{y-2}=2\\)
\n\\(\\frac{6}{x-1}-\\frac{3}{y-2}=1\\)
\nLet \\(\\frac{1}{x-1}=p \\text { and } \\frac{1}{y-2}=q\\)
\n\\(5\\left(\\frac{1}{x-1}\\right)+\\frac{1}{y-2}=2\\) ……………. (i)
\n\\(6\\left(\\frac{1}{x-1}\\right)-3\\left(\\frac{1}{y-2}\\right)=1\\) …………… (ii)
\n5p + 1 = 2 …………… (iii)
\n6p – 3q = 1 …………… (iv)
\nBy solving equation \\(\\mathrm{p}=\\frac{1}{3}, \\quad \\mathrm{q}=\\frac{1}{3}\\)
\nSubstituting \\(\\frac{1}{x-1}\\) for p,
\n\\(p=\\frac{1}{x-1}=\\frac{1}{3}\\)
\nx – 1 = 3
\n\u2234 x = 4
\nSubstituting \\(\\frac{1}{y-2}\\) for q,
\n\\(q=\\frac{1}{y-2}=\\frac{1}{3}\\)
\ny – 2 = 3
\n\u2234 y = 5
\n\u2234 x = 4, y = 5<\/p>\n

(v) \\(\\frac{7 x-2 y}{x y}=5\\) ……………. (i)
\n\\(\\frac{8 x+7 y}{x y}=15\\) ………………… (ii)
\nFrom eqn. (i)
\n\\(\\frac{7 x-2 y}{x y}=5\\)
\n7x – 2y = 5xy ………….. (i)
\nFrom eqn. (ii)
\n\\(\\frac{8 x+7 y}{x y}=15\\)
\n8x + 7y = 15xy ……………… (ii)
\nEqm (i) \u00d7 7 and Eqn. (ii) \u00d7 2, we have
\n49x – 14y = 35xy ………….. (iii)
\n16x + 14y = 30xy ……………. (iv)
\nFrom eqn. (iii) + eqn. (iv)
\n\"KSEEB
\n\u2234 \\(\\frac{65 \\mathrm{x}}{65 \\mathrm{x}}=\\mathrm{y}\\)
\n1 = y
\n\u2234 y = 1
\nSubstituting the value of ‘y’ in eqn. (i)
\n7x – 2y = 5xy
\n7x – 2 (1) = 5 \u00d7 x \u00d7 1
\n7x – 2 = 5x
\n7x – 5x = 2
\n2x = 2
\n\u2234 x = 1
\n\u2234 x = 1, y = 1<\/p>\n

(vi) 6x + 3y = 6xy ………………. (i)
\n2x + 4y = 5xy ………………… (ii)
\nMultiplying eqn. (ii) by 3
\n3(2x + 4y = 5xy)
\n6x + 12y = 18xy ………………. (iii)
\nFrom eqn. (i) – eqn. (ii)
\n\"KSEEB
\n9y = 9xy
\n\\(\\frac{9 y}{9 y}=x\\)
\n1 = x
\n\u2234 x = 1
\nSubstituting the value of ‘x’ in eqn (i),
\n6x + 3y = 6xy
\n6 \u00d7 1 + 3y = 6 \u00d7 1 \u00d7 y
\n6 + 3y = 6y
\n6y – 3y = 6
\n3y = 6
\n\u2234 \\(\\quad y=\\frac{6}{3}\\)
\n\u2234 y = 2
\n\u2234 x = 1, y = 2<\/p>\n

(viii) \\(\\frac{10}{x+y}+\\frac{2}{x-y}=4\\)
\n\\(\\frac{15}{x+y}+\\frac{5}{x-y}=-2\\)
\nLet \\(\\mathrm{u}=\\frac{1}{\\mathrm{x}+\\mathrm{y}}, \\mathrm{v}=\\frac{1}{\\mathrm{x}-\\mathrm{y}}\\) then,
\n10u + 2v = 4 …………. (i)
\n15u – 5v = -2 …………. (ii)
\nMultiplying eqn. (i) by 5 and multiplying eqn. (ii) by 2 and then by adding
\n\"KSEEB
\n\u2234 \\(\\quad \\mathbf{u}=\\frac{1}{5}\\)
\nSubstituting the value of ‘u’ in eqn. (i)
\n\\(10 \\times \\frac{1}{5}+2 v=4\\)
\n2 + 2v = 4
\n2v = 4 – 2
\n2v = 2
\n\u2234 v = 1
\n\u2234 u = 1\/5 , v =1
\n\\(\\frac{1}{x+y}=\\frac{1}{5} \\quad \\frac{1}{x-y}=1\\)
\n\u2234 x + y = 5 x – y = 1
\nx + y = 5 …………….. (iii)
\nx – y = 1 …………….. (iv)
\nFrom eqn. (iii) + eqn. (iv)
\n\"KSEEB
\nx = 3
\n\u2234 Substituting the value of ‘x’ in (iii)
\nx + y = 5
\n3 + y = 5
\ny = 5 – 3
\ny = 2
\n\u2234 x = 3, y = 2<\/p>\n

(viii) \\(\\frac{1}{3 x+y}+\\frac{1}{3 x-y}=\\frac{3}{4}\\) ………….. (i)
\n\\(\\frac{1}{2(3 x+y)}-\\frac{1}{2(3 x-y)}=\\frac{-1}{8}\\) ……………. (ii)
\n\"KSEEB
\nSubstituting the value of ‘p’ in eqn. (iii)
\n\\(p+q=\\frac{3}{4}\\)
\n\\(\\frac{1}{4}+q=\\frac{3}{4}\\)
\n\u2234 \\(\\quad q=\\frac{3}{4}-\\frac{1}{4}\\)
\n\u2234 \\(\\quad q=\\frac{1}{2}\\)
\n\u2234 \\(\\quad p=\\frac{1}{3 x+y}=\\frac{1}{4}\\)
\n3x + y = 4 ……………. (v)
\n\\(q=\\frac{1}{3 x-y}=\\frac{1}{2}\\)
\n3x – y = 2 ……………… (vi)
\nFrom eqn. (v) + eqn. (vi)
\n3x + y = 4
\n+3x – y = 2
\n6x = 6
\n\u2234 \\(\\quad x=\\frac{6}{6}=1\\)
\nSubstituting the value of ‘x’ in eqn. (v)
\n3x + y = 4
\n3 \u00d7 1 + y = 4
\n3 + y = 4
\n\u2234 y = 4 – 3
\n\u2234 y = 1
\n\u2234 x = 1, y = 1<\/p>\n

Question 2.
\nFormulate the following problems as a pair of equations, and hence find their solutions :
\n(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
\nSolution:
\nLet speed of Ritu
\nin still water be x km\/hr.
\nin current water be y km\/hr
\nSpeed of water while
\nupstream is (x + y) km\/hr,
\ndownstream is (x – y) km\/hr,
\n2(x + y) = 20
\nx + y = 10 ………… (i)
\n2(x – y) = 4
\nx – y = 2 …………… (ii)
\nBy Adding eqn. (i) and eqn. (ii),
\n\"KSEEB
\n\u2234 \\(\\quad x=\\frac{12}{2}=6\\)
\nSubstituting the value, x = 6 in eqn. (i),
\nx + y = 10
\n6 + y = 10
\ny = 10 – 6
\n\u2234 y = 4
\n\u2234 Speed of Ritu
\nin still water, x = 6 km\/hr.
\nin current water, y = 4 km\/hr.<\/p>\n

(ii) 2 women and 5 men can together finish an embroidery work in 4 days,, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. ‘
\nSolution:
\nNumber of days required
\nfor women be \u2019x’
\nfor men be ‘y’
\nWork done by woman in 1 day \\(=\\frac{1}{x}\\)
\nWork done by man in 1 day \\(=\\frac{1}{y}\\)
\n\\(4\\left(\\frac{2}{x}+\\frac{5}{4}\\right)=1\\)
\n\\(\\frac{2}{x}+\\frac{5}{4}=\\frac{1}{4}\\) ………….. (i)
\n\\(3\\left(\\frac{3}{x}+\\frac{6}{y}\\right)=1\\)
\n\\(\\frac{3}{x}+\\frac{6}{y}=\\frac{1}{3}\\) ……………… (ii)
\nLet \\(\\frac{1}{{x}}={p} \\text { and } \\frac{1}{y}={q}\\)
\n\u2234 \\(2 p+5 q=\\frac{1}{4}\\)
\n8p + 20q = 1 ……….. (iii)
\n\\(3 p+6 q=\\frac{1}{3}\\)
\n9p + 18q = 1 ……………. (iv)
\nBy Cross-multiplication
\n\\(\\frac{\\mathrm{p}}{-20-(-18)}=\\frac{\\mathrm{q}}{-9-(-8)}=\\frac{1}{144-180}\\)
\n\\(\\frac{p}{-2}=\\frac{q}{-1}=\\frac{1}{-36}\\)
\nIf \\(\\frac{p}{-2}=\\frac{1}{-36}\\) then \\(p=\\frac{-2}{-36}=\\frac{1}{18}\\)
\nIf \\(\\frac{q}{-1}=\\frac{1}{-36}\\) then \\(q=\\frac{-1}{-36}=\\frac{1}{36}\\)
\n\\(p=\\frac{1}{x}=\\frac{1}{18} \\quad \u2234 \\quad x=18\\)
\n\\(q=\\frac{1}{y}=\\frac{1}{36} \\quad \u2234 \\quad y=36\\)
\n\u2234 Number of days taken by one woman to complete work, x = 18
\nNumber of days taken by one man to complete the work, y = 36.<\/p>\n

(iii) Roohl travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
\nSolution:
\nLet speed of the train be x’ km\/hr. and Speed of the bus is ‘y’ km\/hr.
\n\\(\\frac{60}{\\mathrm{x}}+\\frac{240}{\\mathrm{y}}=4\\) …………… (i)
\n\\(\\frac{100}{x}+\\frac{200}{y}=\\frac{25}{6}\\) …………… (ii)
\nIf \\(\\frac{1}{\\mathbf{x}}=\\mathbf{p} \\quad \\frac{1}{\\mathbf{y}}=\\mathbf{q}\\) then
\n60p + 240q = 4 ……………… (iii)
\n\\(100 p+200 q=\\frac{25}{6}\\) ……………….. (iv)
\nMultiplying eqn. (iii) by 10, Mulitplying eqn. (iv) by 6 and then eqn. (iii) – eqn. (iv), we have
\n\"KSEEB
\n\u2234 \\(\\quad q=\\frac{15}{1200}\\)
\n\u2234 \\(\\quad q=\\frac{1}{80}\\)
\nSubstituting the value of \\(q=\\frac{1}{80}\\) in eqn. (iii)
\n\\(60 p+240 \\times \\frac{1}{80}=4\\)
\n60p + 3 = 4
\n60p =4 – 3
\n\u2234 \\(\\mathrm{p}=\\frac{1}{60}\\)
\n\u2234 \\(\\quad \\mathrm{p}=\\frac{1}{60}=\\frac{1}{\\mathrm{x}}\\)
\n\u2234 x = 60 km\/hr.
\n\u2234 \\(\\quad q=\\frac{1}{80}=\\frac{1}{y}\\)
\n\u2234 y = 80 km\/hr.
\n\u2234 Speed of train, x = 60 km\/hr.
\nSpeed of bus, y = 80 km\/hr.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 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 3.6, 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 3.6\u00a0are 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 3.6. Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair …<\/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\/750"}],"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=750"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/750\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=750"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=750"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=750"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}