{"id":711,"date":"2020-09-15T06:29:51","date_gmt":"2020-09-15T00:59:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=711"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-5\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5<\/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 3.5.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5<\/h2>\n

Question 1.
\nWhich of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
\n(i) x – 3y – 3 = 0
\n3x – 9y – 2 = 0
\n(ii) 2x + y = 5
\n3x + 2y = 8
\n(iii) 3x – 5y = 20
\n6x – 10y = 40
\n(iv) x – 3y – 7 = 0
\n3x – 3y – 15 = 0
\nSolution:
\n(i) x – 3y – 3 = 0 …………. (i)
\n3x – 9y – 2 = 0 ……………… (ii)
\nHere, a1<\/sub> = 1, b1<\/sub> = -3, c1<\/sub> = -3
\na2<\/sub> = 3, b2<\/sub> = -9, c2<\/sub> = -2
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{1}{3} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{-3}{-9}=\\frac{1}{3}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-3}{-2}=\\frac{3}{2}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{2}}{c_{2}}\\)
\n\u2234 No solution for these.<\/p>\n

(ii) 2x + y = 5
\n3x + 2y = 8
\n2x + y – 5 = 0 \u2192 (1)
\n3x + 2y – 8 = 0 \u2192 (2)
\nHere, a1<\/sub>\u00a0= 2, b1<\/sub>\u00a0= 1, c1<\/sub>= – 5
\na2<\/sub>\u00a0= 3, b2<\/sub>\u00a0= 2, c2<\/sub> = – 8
\n\"\\frac{a_{1}}{a_{2}}=\\frac{2}{3}
\nHere,\u00a0\"\\frac{a_{1}}{a_{2}}
\n\u2234 Many solutions are there for these
\n\"KSEEB
\n\\(\\frac{\\mathbf{x}}{1 x-8-(2 x-5)}=\\frac{y}{-5 \\times 3-(-8 \\times 2)}=\\frac{1}{2 \\times 2-3 \\times 1}\\)
\n\\(\\frac{x}{-8+10}=\\frac{y}{-15+16}=\\frac{1}{4-3} \\)
\n\\(\\frac{\\mathbf{x}}{+2}=\\frac{\\mathbf{y}}{1}=\\frac{1}{1}\\)
\n\\(\\frac{\\mathbf{x}}{+2}=\\frac{1}{1} \\quad \\mathbf{x}=2\\)
\n\\(\\frac{y}{1}=\\frac{1}{1} \\quad y=1\\)
\n\u2234 Unique solution: x = 2, y = 1<\/p>\n

(iii) 3x – 5y = 20
\n6x – 10y = 40
\n3x – 5y – 20 = 0 \u2192 (1)
\n6x – 10y – 40 = 0 \u2192 (2)
\nHere, a1<\/sub>\u00a0= 3 b1<\/sub> = – 5, c1<\/sub> = – 20
\na2<\/sub>\u00a0= 6, b2\u00a0<\/sub>= – 10, c2<\/sub> = – 40
\n\"\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{3}{6}=\\frac{1}{2}
\n\"\\frac{c_{1}}{c_{2}}=\\frac{-20}{-30}=\\frac{1}{2}\"
\nHere,\u00a0\"\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\"
\nInfinite solutions are there for these.<\/p>\n

(iv) x – 3y – 7 = 0 …………… (i)
\n3x – 3y – 15 = 0 …………….. (ii)
\na1<\/sub> = 1, b1<\/sub> = -3, c1<\/sub> = -7
\na2<\/sub> = 3, b2<\/sub> = -3, c2<\/sub> = -15
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{1}{3} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-3}{-3}=\\frac{1}{1}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-7}{-15}=\\frac{7}{15}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}} \\neq \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}\\)
\n\u2234 Many solutions are there
\n\"KSEEB
\n\\(\\frac{x}{-3 x-15-(-3 x-7)}=\\frac{y}{-7 \\times 3-(-15 \\times 1)}=\\frac{1}{1 x-3-(3 x-3)}\\)
\n\\(\\frac{x}{45-21}=\\frac{y}{-21+15}=\\frac{1}{-3+9}\\)
\n\\(\\frac{x}{24}=\\frac{y}{-6}=\\frac{1}{6}\\)
\nIf \\(\\frac{x}{24}=\\frac{1}{6}\\) then
\n6x = 24
\n\\(x=\\frac{24}{6}=4\\)
\nIf \\(\\frac{y}{-6}=\\frac{1}{6}\\) then
\n6y = -6
\n\\(y=\\frac{-6}{6}\\)
\n\u2234 y = -1
\n\u2234 Unique solutions are there for these x = 4, y = -1.<\/p>\n

Question 2.
\n(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
\n2x + 3y = 7
\n(a – b) x + (a + b) y = 3a + b – 2
\n(ii) For which value of k will the following pair of linear equations have no solution ?
\n3x + y = 1
\n(2k – 1)x + (k – 1)y = 2k + 1
\nSolution:
\n(i) 2x + 3y = 7
\n2x + 3y – 7 = 0 ………… (i)
\n(a – b)x + (a + b)y = (3a + b – 2) ………… (ii)
\n(a – b)x + (a + b)y – (3a + b – 2) ……………. (ii)
\nHere, a1<\/sub> =2, b1<\/sub> = 3, c1<\/sub> = -7
\na2<\/sub> = (a – b), b2<\/sub> = (a + b), c2<\/sub> = -(3a + b – 2)
\nIf it has infinite solutions, then we have
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\)
\n\\(\\frac{2}{(a-b)}=\\frac{3}{(a+b)}=\\frac{-7}{-(3 a+b-2)}\\)
\nIf \\(\\frac{2}{(a-b)}=\\frac{-7}{(-3 a-b+2)}\\) then
\n2(-3a – b + 2) = -7(a – b)
\n-6a – 2b + 4 = -7a + 7b
\n-6a + 7a – 2b – 7b = -4
\na – 9b = 4 ………… (i)
\nIf \\(\\frac{3}{(a+b)}=\\frac{-7}{(-3 a-b+2)}\\) then
\n3(-3a – b + 2) = -7(a + b)
\n-9a- 3b + 6 = -7a – 7b
\n-9a + 7a – 3b + 7b = -6
\n-2a + 4b = -6
\n-a + 2b = -3 …………. (ii)
\nFrom eqn. (i) + eqn. (ii),
\n\"KSEEB
\n7b = 7
\n\\(b=\\frac{7}{7}=1\\)
\nSubstituting the value of ‘b’ in eqn. (i),
\na – 9b = -4
\na – 9(1) = -4
\na – 9 = -4
\na = -4 + 9
\na = 5
\n\u2234 a = 5, b= 1.<\/p>\n

(ii) If the linear pair do not have solutions, then we have
\n\"\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}
\n3x + y – 1 = 0 \u2192 (1)
\n(2k – 1)x + (k – 1)y – (2k + 1) = 0 \u2192 (2)
\nHere a1<\/sub>\u00a0= 3, b1<\/sub>\u00a0= 1, c1<\/sub>= 1
\na2<\/sub> = (2k – 1), b2<\/sub> = (k – 1), c2<\/sub>\u00a0= -(2k + 1)
\n\"\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}=\\frac{\\mathbf{b}_{\\mathbf{l}}}{\\mathbf{b}_{2}}\"
\n\"\\frac{3}{(2
\n3(k – 1) = 1(2k – 1)
\n3k – 3 = 2k – 2
\n3k – 2k = – 2 + 3
\n\u2234 k = 1.<\/p>\n

Question 3.
\nSolve the following pair of linear equations by the substitution and cross-multiplication methods:
\n8x + 5y = 9
\n3x + 2y = 4
\nSolution:
\n8x + 5y = 9 ……………. (i)
\n3x + 2y = 4 …………….. (ii)
\n(1) Substitution Method:
\n8x + 5y = 9
\n8x = 9 – 5y
\n\\(x=\\frac{9-5 y}{8}\\)
\nSubstituting the value of ‘x’ in eqn (i)
\n3x + 2y = 4
\n\\(3\\left(\\frac{9-5 y}{8}\\right)+2 y=4\\)
\n\\(\\frac{27-15 y}{8}+\\frac{2 y}{1}=4\\)
\n\\(\\frac{27-15 y+164}{8}=4\\)
\n27 + y = 8 \u00d7 4
\n27 + y = 32
\ny = 32 – 27
\n\u2234 y = 5
\nSubstituting the value of ‘y’ in
\n\\(x=\\frac{9-5 y}{8}\\)
\n\\(=\\frac{9-5 \\times 5}{8}\\)
\n\\(=\\frac{9-25}{8}\\)
\n\\(=\\frac{-16}{8}\\)
\n\u2234 x = -2
\n\u2234 x = -2, y = 5
\n(2) Cross-Multiplication Method:
\n8x + 5y – 9 = 0 …………. (i)
\n3x + 2y – 4 = 0 …………. (ii)
\nHere, a1<\/sub> = 8 b1<\/sub> = 5 c1<\/sub>=-9
\na2<\/sub> = 3, b2<\/sub> = 2, c2<\/sub> = -4
\n8x + 5y = 9 3x + 2y = 4
\n\"KSEEB
\n\\(\\frac{x}{5 x-4-(2 x-9)}=\\frac{y}{-9 \\times 3-(-4 \\times 8)}=\\frac{1}{8 \\times 2-3 \\times 5}\\)
\n\\(\\frac{x}{-20+18}=\\frac{y}{-27+32}=\\frac{1}{16-15}\\)
\n\\(\\frac{\\mathbf{x}}{-2}=\\frac{y}{5}=\\frac{1}{1}\\)
\n\\(\\frac{\\mathrm{x}}{-2}=1 \\quad \\mathrm{x}=-2\\)
\n\\(\\frac{\\mathrm{y}}{5}=1 \\quad \\mathrm{y}=5\\)<\/p>\n

Question 4.
\nForm the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
\n(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
\nSolution:
\nFixed hostel charges are Rs. x.
\nAmount to be paid to be Rs. y.
\nx + 20y = 1000 …………. (i)
\nx + 26y = 1180 …………. (ii)
\nBy elimination method, we can find solution.
\nSubtracting eqn. (ii) from eqn. (i),
\n\"KSEEB
\n– 6y =- 180
\n\\(\\mathbf{y}=\\frac{180}{6}\\)
\n\u2234 y = Rs. 30
\nSubstituting the value of ‘y’ in eqn. (i),
\nx + 20y = 1000
\nx + 20(30) = 1000
\nx + 600 = 1000
\n\u2234 x = 1000 – 600
\n\u2234 x = Rs. 400
\nFixed charge, x = Rs. 400
\nAmount to be paid, y = Rs. 30.<\/p>\n

(ii) A fraction becomes \\(\\frac{1}{3}\\) when 1 is subtracted from the numerator and it becomes \\(\\frac{1}{4}\\) when 8 is added to its 4 denominator. Find the fraction.
\nSolution:
\nLet the fraction be\u00a0\"\\frac{x}{y}\"
\n\"\\frac{x-1}{y}=\\frac{1}{3}\"
\n3(x – 1) = y
\n3x – 3 = y
\n3x – y – 3 = 0 \u2192 (1)
\n\"\\frac{x}{y+8}=\\frac{1}{4}\"
\n4x = y + 8
\n4x – y – 8 = 0 \u2192 (2)
\nBy Substitution method we can find out solution:
\nFrom eqn. (i),
\n3x – y – 3 = 0
\ny = 3x – 3
\nSubstituting the value of y\u2019 in eqn. (2),
\n4x – y – 8 = 0
\n4x – (3x – 3) – 8 = 0
\n4x – 3x + 3 – 8 = 0
\nx – 5 = 0
\n\u2234 x = 5.
\nSubstituting the value of \u2018x\u2019 in
\ny = 3x – 3,
\n= 3(5) – 3
\n= 15 – 3
\n\u2234 y = 12
\n\u2234 x = 5, y = 12
\n\u2234 Fraction is\u00a0\"\\frac{x}{y}=\\frac{5}{12}\"<\/p>\n

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
\nSolution:
\nLet the number of right answers be ‘x’, and the number of wrong answers be y’.
\n3x – y = 40 ……….. (i)
\n4x – 2y = 50 ………. (ii)
\nBy Elimination method we can find our solution :
\n3x – y = 40 ……… (i)
\n2x – y = 25 ……….. (ii)
\nSubtracting eqn. (ii) from eqn. (i),
\n\"KSEEB
\nSubstituting the value of ‘x’ in eqn. (i),
\n3x – y = 40
\n3(15) – y = 40
\n45 – y = 40
\n-y = 40 – 45
\n-y = -5
\n\u2234 y = 5
\nQuestions of right answers, x = 15
\nQuestions of wrong answers, y = 5
\n\u2234 Total number of questions = x + y = 15 + 5 = 20<\/p>\n

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ?
\nSolution:
\nDistance between places A and B is 100 km.
\nSpeed of A car us ‘u’ km\/hr.
\nSpeed of B car is ‘v’ km\/hr.
\nIf the cars travel in the same direction at different speeds they meet in 5 hours about 20 km.
\n\u2234 u – v = 20 ………. (i)
\nIf the cars travel in the same direction, total distance covered =100 km.
\n\u2234 u + v = 100 …….. (ii)
\nWe can solve by Elimination method:
\nFrom eqn. (i) + eqn. (ii),
\n\"KSEEB
\n\\(u=\\frac{120}{2}\\)
\n\u2234 u = 60 km\/hr.
\nSubstituting the value of ‘u’ in Eqn. (ii),
\nu + v = 100
\n60 + v = 100
\nv = 100 – 60
\nv = 40 km\/hr.
\n\u2234 u = 60 km\/hr.
\nv = 40 km\/hr.<\/p>\n

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
\nSolution:
\nLength of rectangle is ‘x’ unit.
\nBreadth of rectangle is ‘y’ unit.
\nArea of rectangle = length \u00d7 breadth = xy
\n(x – 5) (y + 3) = xy – 9
\nxy + 3x – 5y – 15 = xy – 9
\n3x – 5y = -9 + 15
\n3x – 5y = 6 …………. (i)
\n(x + 3) (y + 2) = xy + 67
\nxy + 2x + 3y + 6 = xy + 67
\n2x + 3y = 67 – 6
\n2x + 3y = 61 ……….. (ii)
\nSolving the equation by Elimination method :
\nMultiplying eqn. (i) by and eqn. (ii) by 5
\n9x – 15y = 18 ………….. (iii)
\n10x + 15y = 305 ………… (iv)
\nFrom eqn. (iii) + eqn. (iv)
\n\"KSEEB
\n\\(x=\\frac{323}{19}\\)
\n\u2234 x = 17 unit.
\nSubstituting the value of ‘x’ in eqn. (i),
\n3x – 5y = 6
\n3(17) – 5y = 6
\n51 – 5y = 6
\n-5y = 6 – 51
\n– 5y = – 45
\n5y = 45
\n\\(y=\\frac{45}{5}\\)
\n\u2234 y = 9 units.
\n\u2234 Length of that rectangle, x = 17 unit
\nbreadth, y = 9 unit.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 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 3.5, 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 3.5\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 3.5. 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\/711"}],"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=711"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/711\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=711"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=711"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}