{"id":4740,"date":"2021-03-02T11:09:59","date_gmt":"2021-03-02T05:39:59","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4740"},"modified":"2021-09-15T11:40:36","modified_gmt":"2021-09-15T06:10:36","slug":"ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-3\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3<\/h2>\n

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers<\/h3>\n

Question 1.
\nSolve each of the following pairs of equations by reducing them to a pair of linear equations.
\ni) \\(\\frac{5}{x-1}\\) + \\(\\frac{1}{y-2}\\) = 2
\n\\(\\frac{6}{x-1}\\) + \\(\\frac{3}{y-2}\\) = 1
\nAnswer:
\nGiven
\n\\(\\frac{5}{x-1}\\) + \\(\\frac{1}{y-2}\\) = 2
\n\\(\\frac{6}{x-1}\\) + \\(\\frac{3}{y-2}\\) = 1
\nPut \\(\\frac{1}{x-1}\\) = a and \\(\\frac{1}{y-2}\\) = b,
\nthen the given equations reduce to
\n5a + b = 2 ……… (1)
\n6a – 3b = 1 ………. (2)
\n\"AP
\n\u21d2 b = \\(\\frac{7}{21}\\) = \\(\\frac{1}{3}\\)
\nSubstituting b = \\(\\frac{1}{3}\\) in equation (1) we get
\n\"AP
\n\u21d2 (x – 1) . 1 = 3 \u00d7 1
\n\u21d2 x – 1 = 3
\n\u21d2 x = 3 + 1 = 4
\nb = \\(\\frac{1}{y-2}\\) \u21d2 \\(\\frac{1}{3}\\) = \\(\\frac{1}{y-2}\\)
\n\u21d2 (y – 2) . 1 = 3 \u00d7 1
\n\u21d2 y – 2 = 3
\n\u21d2 y = 3 + 2 = 5
\n\u2234 Solution (x, y) = (4, 5)<\/p>\n

\"AP<\/p>\n

ii) \\(\\frac{x+y}{xy}\\) = 2;
\n\\(\\frac{x-y}{xy}\\) = 6
\nAnswer:
\nGiven
\n\\(\\frac{x+y}{xy}\\) = 2
\n\u21d2 \\(\\frac{x}{xy}\\) + \\(\\frac{y}{xy}\\) = 2
\n\u21d2 \\(\\frac{1}{y}\\) + \\(\\frac{1}{x}\\) = 2
\n\\(\\frac{x-y}{xy}\\) = 6
\n\u21d2 \\(\\frac{x}{xy}\\) – \\(\\frac{y}{xy}\\) = 6
\n\u21d2 \\(\\frac{1}{y}\\) – \\(\\frac{1}{x}\\) = 6
\nTake \\(\\frac{1}{x}\\) = a and \\(\\frac{1}{y}\\) = b,
\nthen the given equations reduces to
\n\"AP
\n\u21d2 b = \\(\\frac{8}{2}\\) = 4
\nSubstituting b = 4 in equation (1) we get
\na + 4 = 2 \u21d2 a = 2 – 4 = -2
\nbut a = \\(\\frac{1}{x}\\) = -2 \u21d2 x = \\(\\frac{-1}{2}\\)
\nb = \\(\\frac{1}{y}\\) = 4 \u21d2 y = \\(\\frac{1}{4}\\)
\n\u2234 Solution (x, y) = \\(\\left(\\frac{-1}{2}, \\frac{1}{4}\\right)\\)<\/p>\n

iii) \\(\\frac{2}{\\sqrt{x}}\\) + \\(\\frac{3}{\\sqrt{y}}\\) = 2;
\n\\(\\frac{4}{\\sqrt{x}}\\) – \\(\\frac{9}{\\sqrt{y}}\\) = -1
\nAnswer:
\nGiven
\n\\(\\frac{2}{\\sqrt{x}}\\) + \\(\\frac{3}{\\sqrt{y}}\\) = 2 and \\(\\frac{4}{\\sqrt{x}}\\) – \\(\\frac{9}{\\sqrt{y}}\\) = -1
\nTake \\(\\frac{1}{\\sqrt{x}}\\) = a and \\(\\frac{1}{\\sqrt{y}}\\) = b,
\nthen the given equations reduces to
\n2a + 3b = 2 …….. (1)
\n4a – 9b = – 1 …….. (2)
\n\"AP
\n\u21d2 b = \\(\\frac{5}{15}\\) = \\(\\frac{1}{3}\\)
\nSubstituting b = \\(\\frac{1}{3}\\) in equation (1) we get
\n2a + 3\\(\\left(\\frac{1}{3}\\right)\\) = 2
\n\u21d2 2a + 1 = 2 \u21d2 2a = 2 – 1 \u21d2 a = \\(\\frac{1}{2}\\)
\n\"AP
\n\u2234 Solution (x, y) = (4, 9)<\/p>\n

\"AP<\/p>\n

iv) 6x + 3y = 6xy
\n2x + 4y = 5xy
\nAnswer:
\nGiven
\n6x + 3y = 6xy
\n\u21d2 \\(\\frac{6x+3y}{xy}\\) = 6
\n\u21d2 \\(\\frac{6x}{xy}\\) + \\(\\frac{3y}{xy}\\) = 6
\n\u21d2 \\(\\frac{6}{y}\\) + \\(\\frac{3}{x}\\) = 6
\n2x + 4y = 5xy
\n\u21d2 \\(\\frac{2x+4y}{xy}\\) = 5
\n\u21d2 \\(\\frac{2x}{xy}\\) + \\(\\frac{4y}{xy}\\) = 6
\n\u21d2 \\(\\frac{2}{y}\\) + \\(\\frac{4}{x}\\) = 6
\nTake \\(\\frac{1}{x}\\) = a and \\(\\frac{1}{y}\\) = b,
\nthen the given equations reduces to
\n3a + 6b = 6 ……. (1)
\n4a + 2b = 5 ……. (2)
\n\"AP
\n\u21d2 b = \\(\\frac{9}{18}\\) = \\(\\frac{1}{2}\\)
\nSubstituting b = \\(\\frac{1}{2}\\) in equation (1) we get
\n3a +6\\(\\left(\\frac{1}{2}\\right)\\) = 6
\n\u21d2 3a = 6 – 3
\n\u21d2 a = \\(\\frac{3}{3}\\) = 1
\nbut a = \\(\\frac{1}{x}\\) = 1 \u21d2 x = 1
\nb = \\(\\frac{1}{y}\\) = \\(\\frac{1}{2}\\) \u21d2 y = 2
\n\u2234 Solution (x, y) = (1, 2)<\/p>\n

v) \\(\\frac{5}{x+y}\\) – \\(\\frac{2}{x-y}\\) = -1
\n\\(\\frac{15}{x+y}\\) + \\(\\frac{7}{x-y}\\) = 10
\nwhere x \u2260 0, y \u2260 0
\nAnswer:
\nGiven
\n\\(\\frac{5}{x+y}\\) – \\(\\frac{2}{x-y}\\) = -1 and
\n\\(\\frac{15}{x+y}\\) + \\(\\frac{7}{x-y}\\) = 10
\nTake \\(\\frac{1}{x+y}\\) = a and \\(\\frac{1}{x-y}\\) = b, then
\nthe given equations reduce to
\n5a – 2b = – 1 ……… (1)
\n15a + 7b = 10 ……… (2)
\n\"AP
\n\u21d2 b = \\(\\frac{-13}{-13}\\) = 1
\nSubstituting b = 1 in equation (1) we get
\n5a – 2(1) = -1
\n\u21d2 5a = -1 + 2
\n\u21d2 5a = 1
\n\u21d2 a = \\(\\frac{1}{5}\\)
\nbut a = \\(\\frac{1}{x+y}\\) = \\(\\frac{1}{5}\\) \u21d2 x + y = 5
\nb = \\(\\frac{1}{x-y}\\) = 1 \u21d2 x – y = 1
\n\u21d2 x = \\(\\frac{6}{2}\\) = 3
\nSolving the above equations
\nSubstituting x = 3 in x + y = 5 we get
\n3 + y = 5 \u21d2 y = 5 – 3 = 2
\n\u2234 Solution (x, y) = (3, 2)<\/p>\n

\"AP<\/p>\n

vi) \\(\\frac{2}{x}\\) + \\(\\frac{3}{y}\\) = 13
\n\\(\\frac{5}{x}\\) – \\(\\frac{4}{y}\\) = -2
\nwhere x \u2260 0, y \u2260 0
\nAnswer:
\nGiven
\n\\(\\frac{2}{x}\\) + \\(\\frac{3}{y}\\) = 13 and
\n\\(\\frac{5}{x}\\) – \\(\\frac{4}{y}\\) = -2
\nTake \\(\\frac{1}{x}\\) = a and \\(\\frac{1}{y}\\) = b, then
\nthe given equations reduce to
\n2a + 3b = 13 ……… (1)
\n5a – 4b = -2 ……… (2)
\n\"AP
\n\u21d2 b = \\(\\frac{69}{23}\\) = 3
\nSubstituting b = 3 in equation (1) we get
\n2a + 3 (3) = 13
\n\u21d2 2a = 13 – 9
\n\u21d2 a = \\(\\frac{4}{2}\\) = 2
\nbut a = \\(\\frac{1}{x}\\) = 2 \u21d2 x = \\(\\frac{1}{2}\\)
\nb = \\(\\frac{1}{y}\\) = 3 \u21d2 y = \\(\\frac{1}{3}\\)
\n\u2234 Solution (x, y) = (\\(\\frac{1}{2}\\), \\(\\frac{1}{3}\\))<\/p>\n

vii) \\(\\frac{10}{x+y}\\) + \\(\\frac{2}{x-y}\\) = 4
\n\\(\\frac{15}{x+y}\\) – \\(\\frac{5}{x-y}\\) = -2
\nAnswer:
\nGiven
\n\\(\\frac{10}{x+y}\\) + \\(\\frac{2}{x-y}\\) = 4 and
\n\\(\\frac{15}{x+y}\\) – \\(\\frac{5}{x-y}\\) = -2
\nTake \\(\\frac{1}{x+y}\\) = a and \\(\\frac{1}{x-y}\\) = b, then
\nthe given equations reduce to
\n10a + 2b = 4 ……… (1)
\n15a – 5b = – 2 ……… (2)
\n\"AP
\n\u21d2 b = \\(\\frac{16}{16}\\) = 1
\nSubstituting b = 1 in equation (1) we get
\n10a + 2(1) = 4
\n\u21d2 10a = 4 – 2
\n\u21d2 a = \\(\\frac{2}{10}\\) = \\(\\frac{1}{5}\\)
\nbut a = \\(\\frac{1}{x+y}\\) = \\(\\frac{1}{5}\\) \u21d2 x + y = 5 ……. (3)
\nb = \\(\\frac{1}{x-y}\\) = 1 \u21d2 x – y = 1 …….. (4)
\nAdding (3) and (4)
\n\"AP
\n\u21d2 x = \\(\\frac{6}{2}\\) = 3
\nSubstituting x = 3 in x + y = 5 we get
\n3 + y = 5 \u21d2 y = 5 – 3 = 2
\n\u2234 Solution (x, y) = (3, 2)<\/p>\n

\"AP<\/p>\n

viii) \\(\\frac{1}{3x+y}\\) + \\(\\frac{1}{3x-y}\\) = \\(\\frac{3}{4}\\)
\n\\(\\frac{1}{2(3x+y)}\\) – \\(\\frac{1}{2(3x-y)}\\) = \\(\\frac{-1}{8}\\)
\nAnswer:
\nGiven
\n\\(\\frac{1}{3x+y}\\) + \\(\\frac{1}{3x-y}\\) = \\(\\frac{3}{4}\\) and
\n\\(\\frac{1}{2(3x+y)}\\) – \\(\\frac{1}{2(3x-y)}\\) = \\(\\frac{-1}{8}\\)
\nTake \\(\\frac{1}{3x+y}\\) = a and \\(\\frac{1}{3x-y}\\) = b, then
\nthe given equations reduce to
\n\"AP
\n\u21d2 a = \\(\\frac{2}{8}\\) = \\(\\frac{1}{4}\\)
\nSubstituting a = \\(\\frac{1}{4}\\) in equation (1) we get
\n\"AP
\nSolving (3) and (4)
\n\"AP
\n\u21d2 x = \\(\\frac{6}{6}\\) = 1
\nSubstituting x = 1 in 3x + y = 4
\n\u21d2 3(1) + y = 4
\n\u21d2 y = 4 – 3 = 1
\n\u2234 The solution (x, y) = (1, 1)<\/p>\n

Question 2.
\nFormulate the following problems as a pair of equations and then find their solutions.
\ni) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
\nAnswer:
\nLet the speed of the boat in still water = x kmph
\nand the speed of the stream = y kmph
\nthen speed in downstream = x + y
\nSpeed in upstream = x – y
\nand time = \\(\\frac{\\text { distance }}{\\text { speed }}\\)
\nBy problem,
\n\\(\\frac{30}{x-y}\\) + \\(\\frac{44}{x+y}\\) = 10
\n\\(\\frac{40}{x-y}\\) + \\(\\frac{55}{x+y}\\) = 13
\nTake \\(\\frac{1}{x-y}\\) = a and \\(\\frac{1}{x+y}\\) = b, then
\nthe given equations reduce to
\n30a + 44b = 10 ……… (1)
\n40a + 55b = 13 ……… (2)
\n\"AP
\n\u21d2 b = \\(\\frac{1}{11}\\)
\nSubstituting b = \\(\\frac{1}{11}\\) in equation (1) we get
\n\"AP
\n\u21d2 x = 8
\nSubstituting x = 8 in x – y = 5 we get
\n8 – y = 5
\n\u21d2 y = 8 – 5 = 3
\n\u2234 The solution (x, y) = (8, 3)
\nSpeed of the boat in still water = 8 kmph
\nSpeed of the stream = 3 kmph.<\/p>\n

\"AP<\/p>\n

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
\nAnswer:
\nLet the speed of the train be x kmph
\nand the speed of the car = y kmph
\nBy problem,
\n\"AP
\nTake \\(\\frac{1}{x}\\) = a and \\(\\frac{1}{y}\\) = b, then
\nthe given equations reduce to
\n15a + 60b = 1 ……… (1)
\n8a + 16b = \\(\\frac{1}{3}\\) \u21d2 24a + 48b = 1 ……… (2)
\n\"AP
\n\u21d2 a = \\(\\frac{-1}{-60}\\) = \\(\\frac{1}{60}\\)
\nSubstituting a = \\(\\frac{1}{60}\\) in equation (1) we get
\n\"AP
\nbut a = \\(\\frac{1}{x}\\) = \\(\\frac{1}{60}\\) \u21d2 x = 60 kmph
\nb = \\(\\frac{1}{y}\\) = \\(\\frac{1}{80}\\) \u21d2 y = 80 kmph
\nSpeed of the train = 60 kmph and
\nspeed of the car = 80 kmph<\/p>\n

iii) 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 and 1 man alone to finish the work.
\nAnswer:
\nLet the time taken by 1 woman to complete the work = x days
\nand time taken by 1 man to complete the work = y days
\n\u2234 Work done by 1 woman in 1 day = \\(\\frac{1}{x}\\)
\nWork done by 1 man in 1 day = \\(\\frac{1}{y}\\)
\nBy problem,
\n\"AP
\nTake \\(\\frac{1}{x}\\) = a and \\(\\frac{1}{y}\\) = b,
\nthen the above equations reduce to
\n2a + 5b = \\(\\frac{1}{4}\\) and 3a + 6b = \\(\\frac{1}{3}\\)
\n\u21d2 8a + 20b = 1 …….. (1) and
\n9a + 18b = 1 ……… (2)
\n\"AP
\n\u21d2 b = \\(\\frac{1}{36}\\)
\nSubstituting b = \\(\\frac{1}{36}\\) in equation (1) we get
\n\"AP
\nbut a = \\(\\frac{1}{x}\\) = \\(\\frac{1}{18}\\) \u21d2 x = 18 and
\nb = \\(\\frac{1}{y}\\) = \\(\\frac{1}{36}\\) \u21d2 y = 36
\n\u2234 Time taken by 1 woman = 18 days
\n1 man = 36 days<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3 10th Class Maths 4th Lesson Pair of Linear Equations in Two …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4740"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=4740"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4740\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4740"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4740"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4740"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}