{"id":6021,"date":"2023-07-23T05:37:54","date_gmt":"2023-07-23T05:37:54","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6021"},"modified":"2023-07-23T05:45:16","modified_gmt":"2023-07-23T05:45:16","slug":"ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 3 (Pair of Linear Equations in Two Variables)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 3 Pair of Linear Equations in Two Variables <\/strong>Exercise 3.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 3 Pair of Linear Equations in Two Variables<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.5<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.6<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-7\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.7<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 3.4\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Solve the following pair of linear equations by the elimination method and the substitution method:<br \/>\n<\/strong><strong>(i) x + y = 5 and 2x \u2013 3y = 4<br \/>\n<\/strong><strong>(ii) 3x + 4y = 10 and 2x \u2013 2y = 2<br \/>\n<\/strong><strong>(iii) 3x \u2013 5y \u2013 4 = 0 and 9x = 2y + 7<br \/>\n<\/strong><strong>(iv) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{x}{2}}\" alt=\"\\mathbf{\\frac{x}{2}}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{2y}{3}}\" alt=\"\\mathbf{\\frac{2y}{3}}\" align=\"absmiddle\" \/> = -1 and x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{y}{3}}\" alt=\"\\mathbf{\\frac{y}{3}}\" align=\"absmiddle\" \/> = 3 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) x + y = 5 and 2x \u2013 3y = 4<br \/>\nBy the method of elimination.<br \/>\n<\/strong>x + y = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (i)<br \/>\n2x \u2013 3y = 4\u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen the equation (i) is multiplied by 2, we get<br \/>\n2x + 2y = 10\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (iii)<br \/>\nWhen equation (ii) is subtracted from (iii), we get<br \/>\n(2x \u2013 3y) &#8211;\u00a0 (2x + 2y) = 4 &#8211; 10<br \/>\n-5y = &#8211; 6<br \/>\n5y = 6<br \/>\ny = 6\/5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (iv)<br \/>\nSubstituting the value of y in eq. (i) we get<br \/>\nx = 5 \u2212 6\/5<br \/>\nx = (25 &#8211; 6)\/5<br \/>\nx = 19\/5<br \/>\n\u2234 x = 19\/5 , y = 6\/5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>By the method of substitution.<br \/>\n<\/strong>From equation (i), we get<br \/>\nx = 5 \u2013 y \u2026\u2026\u2026\u2026.. (v)<br \/>\nWhen the value is put in equation (ii), we get<br \/>\n2(5 \u2013 y) \u2013 3y = 4<br \/>\n-5y = -6<br \/>\ny =\u00a06\/5<br \/>\nWhen the values are substituted in equation (v), we get<br \/>\nx =5\u2212 6\/5<br \/>\nx = (25 &#8211; 6)\/5<br \/>\nx = 19\/5<br \/>\n\u2234 x = 19\/5, y = 6\/5 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>\u00a0<\/strong><strong>(ii) 3x + 4y = 10 and 2x \u2013 2y = 2<br \/>\n<\/strong><strong>By the method of elimination.<br \/>\n<\/strong>3x + 4y = 10\u00a0 \u2026\u2026\u2026\u2026.. (i)<br \/>\n2x \u2013 2y = 2\u00a0 \u00a0\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen the equation (i) and (ii) are multiplied by 2, we get<br \/>\n4x \u2013 4y = 4 \u2026\u2026\u2026\u2026.. (iii)<br \/>\nWhen equations (i) and (iii) are added, we get<br \/>\n(3x + 4y) +\u00a0 (4x \u2013 4y) = 10 + 4<br \/>\n7x = 14<br \/>\nx = 14\/7<br \/>\nx = 2\u00a0 \u2026\u2026\u2026\u2026.. (iv)<br \/>\nSubstituting equation (iv) in (i), we get<br \/>\n6 + 4y = 10<br \/>\n4y = 4<br \/>\ny = 1<br \/>\nHence, x = 2 and y = 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>By the method of substitution<br \/>\n<\/strong>From equation (ii), we get<br \/>\nx = 1 + y\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (v)<br \/>\nSubstituting equation (v) in equation (i), we get<br \/>\n3(1 + y) + 4y = 10<br \/>\n3 + 3y + 4y\u00a0 = 10<br \/>\n7y\u00a0 = 7<br \/>\ny = 1<br \/>\nWhen y = 1 is substituted in equation (v), we get<br \/>\nx = 1 + 1 = 2<br \/>\n<strong>Therefore, x = 2 and y = 1<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 3x \u2013 5y \u2013 4 = 0 and 9x = 2y + 7<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>By the method of elimination.<br \/>\n<\/strong>3x \u2013 5y \u2013 4 = 0 \u2026\u2026\u2026\u2026.. (i)<br \/>\n9x = 2y + 7<br \/>\n9x \u2013 2y \u2013 7 = 0 \u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen the equation (i) is multiplied by 3, we get,<br \/>\n9x \u2013 15y \u2013 12 = 0 \u2026\u2026\u2026\u2026.. (iii)<br \/>\nWhen equation (iii) is subtracted from equation (ii), we get<br \/>\n13y = -5<br \/>\ny = -5\/13\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (iv)<br \/>\nWhen equation (iv) is substituted in equation (i), we get<br \/>\n3x + 25\/13 \u2212 4 = 0<br \/>\n3x = 4 &#8211; 25\/13<br \/>\n3x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(52-25)}{13}\" alt=\"\\frac{(52-25)}{13}\" align=\"absmiddle\" \/><br \/>\n3x = 27\/13<br \/>\nx = 9\/13<br \/>\n<strong>\u2234 x = 9\/13 and y = -5\/13\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>By the method of substitution.<br \/>\n<\/strong>From equation (i), we get<br \/>\nx = (5y + 4)\/3\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (v)<br \/>\nPutting the value (v) in equation (ii), we get<br \/>\n9(5y + 4)\/3 \u2212 2y \u22127=0<br \/>\n13y = -5<br \/>\ny = -5\/13<br \/>\nSubstituting this value in equation (v), we get<br \/>\nx = (5(-5\/13) + 4)\/3<br \/>\nx = 9\/13<br \/>\n\u2234 x = 9\/13, y = -5\/13<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) x\/2 + 2y\/3 = -1 and x &#8211; y\/3 = 3<br \/>\n<\/strong><strong>By the method of elimination.<br \/>\n<\/strong>3x + 4y = -6\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (i)<br \/>\nx\u00a0 &#8211; y\/3 = 3<br \/>\n3x \u2013 y = 9\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen equation (ii) is subtracted from equation (i), we get<br \/>\n5y = -15<br \/>\ny = -3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026.. (iii)<br \/>\nWhen equation (iii) is substituted in (i), we get<br \/>\n3x \u2013 12 = -6<br \/>\n3x = 6<br \/>\nx = 2<br \/>\nHence, x = 2 , y = -3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>By the method of substitution.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">From equation (ii), we get<br \/>\nx = (y + 9)\/3\u00a0 \u00a0 \u2026\u2026\u2026\u2026.. (v)<br \/>\nPutting the value obtained from equation (v) in equation (i), we get<br \/>\n3(y + 9)\/3 + 4y =\u22126<br \/>\n5y = -15<br \/>\ny = -3<br \/>\nWhen y = -3 is substituted in equation (v), we get<br \/>\nx = (-3 + 9)\/3<br \/>\nx = 2<br \/>\nTherefore, x = 2 and y = -3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:<br \/>\n<\/strong><strong>(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the fraction be a\/b<br \/>\nAccording to the given information,<br \/>\n(a + 1)\/(b &#8211; 1) = 1<br \/>\n(a + 1) = (b &#8211; 1)<br \/>\na \u2013 b = -2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\na\/(b+1) = 1\/2<br \/>\n2a = b + 1<br \/>\n2a &#8211; b = 1\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen equation (i) is subtracted from equation (ii), we get<br \/>\na = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (iii)<br \/>\nWhen a = 3 is substituted in equation (i), we get<br \/>\n3 \u2013 b = -2<br \/>\n-b = -5<br \/>\nb = 5<br \/>\nHence, the fraction is 3\/5.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(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?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let us assume the present age of Nuri is x.<br \/>\nAnd the present age of Sonu is y.<br \/>\nAccording to the given condition, we can write as<br \/>\nx \u2013 5 = 3(y \u2013 5)<br \/>\nx \u2013 3y = &#8211; 10\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (1)<br \/>\nNow,<br \/>\nx + 10 = 2(y + 10)<br \/>\nx \u2013 2y = 10\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (2)<br \/>\nSubtract eq. 1 from 2 to get<br \/>\ny = 20\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (3)<br \/>\nSubstituting the value of y in eq.1, we get<br \/>\nx \u2013 3 \u00d7 20 = -10<br \/>\nx \u2013 60 = -10<br \/>\nx = 50<br \/>\nTherefore,<br \/>\nAge of Nuri is 50 years<br \/>\nAge of Sonu is 20 years<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(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.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the unit digit and tens digit of a number be x and y, respectively.<br \/>\nThen, number (n) = 10B + A<br \/>\nN after reversing the order of the digits = 10A + B<br \/>\nAccording to the given information, A + B = 9 \u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\n9(10B + A) = 2(10A + B)<br \/>\n88 B \u2013 11 A = 0<br \/>\n-A + 8B = 0\u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nAdding the equations (i) and (ii), we get<br \/>\n9B = 9<br \/>\nB = 1\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (3)<br \/>\nSubstituting this value of B in equation (i), we get A = 8<br \/>\nHence, the number (N) is 10B + A<br \/>\n= 10 \u00d7 1 + 8<br \/>\n= 18<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) Meena went to a bank to withdraw Rs.2,000. 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.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let the number of Rs.50 notes be A and the number of Rs.100 notes be B.<br \/>\nAccording to the given information,<br \/>\nA + B = 25\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\n50A + 100B = 2000\u00a0 \u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen equation (i) is multiplied by (ii), we get<br \/>\n50A + 50B = 1250\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (iii)<br \/>\nSubtracting equation (iii) from equation (ii), we get<br \/>\n50B = 750<br \/>\nB = 15<br \/>\nSubstituting in equation (i), we get<br \/>\nA = 10<br \/>\nHence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(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 a book she kept for five days. Find the fixed charge and the charge for each extra day.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.<br \/>\nAccording to the information given,<br \/>\nA + 4B = 27\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\nA + 2B = 21\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nWhen equation (ii) is subtracted from equation (i), we get<br \/>\n2B = 6<br \/>\nB = 3\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (iii)<br \/>\nSubstituting B = 3 in equation (i), we get<br \/>\nA + 12 = 27<br \/>\nA = 15<br \/>\nHence, the fixed charge is Rs. 15.<br \/>\nAnd the charge per day is Rs. 3.<\/span><\/p>\n<p><span style=\"font-family: Georgia, Palatino;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 3 (Pair of Linear Equations in Two Variables)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 3 Pair of Linear Equations in Two Variables Exercise 3.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 3 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1058,1062,1044,1049,1048],"class_list":["post-6021","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-3-pair-of-linear-equations-in-two-variables-solutions","tag-ncert-class-10-mathematics-exercise-3-4-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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