{"id":4914,"date":"2016-05-19T09:49:00","date_gmt":"2016-05-19T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-3-4\/"},"modified":"2023-03-22T15:17:06","modified_gmt":"2023-03-22T09:47:06","slug":"ncert-solutions-for-class-10-maths-exercise-3-4","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 3.4"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#NCERT_Solutions_for_Class_10_Maths_Pair_of_Linear_Equations_in_Two_Variables\" >NCERT Solutions for Class 10 Maths\u00a0Pair of Linear Equations in Two Variables<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#Substitution_method\" >Substitution method:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#Substitution_Method\" >Substitution Method:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#Elimination_method\" >Elimination method:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#Substitution_method-2\" >Substitution method:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#2_Form_the_pair_of_linear_equations_in_the_following_problems_and_find_their_solutions_if_they_exist_by_the_elimination_method\" >2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#NCERT_Solutions_for_Class_10_Maths_Exercise_34\" >NCERT Solutions for Class 10 Maths Exercise 3.4<\/a><ul class='ez-toc-list-level-5' ><li class='ez-toc-heading-level-5'><ul class='ez-toc-list-level-5' ><li class='ez-toc-heading-level-5'><ul class='ez-toc-list-level-5' ><li class='ez-toc-heading-level-5'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#_Download_CBSE_Question_Paper_2018\" >\u00a0Download CBSE Question Paper 2018<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-3-4\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 3.4 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Maths\u00a0Pair of Linear Equations in Two Variables\u00a0<a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-pair-of-linear-equations-in-two-variables\/1205\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 3.4\" width=\"187\" height=\"230\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Pair_of_Linear_Equations_in_Two_Variables\"><\/span>NCERT Solutions for Class 10 Maths\u00a0Pair of Linear Equations in Two Variables<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><strong>1. Solve the following pair of linear equations by the elimination method and the substitution method: <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(i) x + y = 5, 2x \u2013 3y = 4<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) 3x + 4y = 10, 2x \u2013 2y = 2<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) 3<em>x <\/em>\u2212 5<em>y <\/em>\u2013 4 = 0, 9<em>x <\/em>= 2<em>y <\/em>+ 7<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) <img decoding=\"async\" style=\"height: 42px; width: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image001.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> x + y = 5 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">2x \u2013 3y = 4 \u2026 (2)<\/p>\n<p style=\"text-align: justify;\"><strong>Elimination method:<\/strong><\/p>\n<p style=\"text-align: justify;\">Multiplying equation (1) by 2, we get equation (3)<\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>+ 2<em>y <\/em>= 10 \u2026 (3)<\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>\u2212 3<em>y <\/em>= 4 \u2026 (2)<\/p>\n<p style=\"text-align: justify;\">Subtracting equation (2) from (3), we get<\/p>\n<p style=\"text-align: justify;\">5<em>y <\/em>= 6\u21d2 <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting value of y in (1), we get<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/>= 5<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= 5 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image003.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image003.png\" \/> and <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/><\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Substitution_method\"><\/span><strong>Substitution method: <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><em>x <\/em>+ <em>y <\/em>= 5 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>\u2212 3<em>y <\/em>= 4 \u2026 (2)<\/p>\n<p style=\"text-align: justify;\">From equation (1), we get,<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>= 5 \u2212 <em>y<\/em><\/p>\n<p style=\"text-align: justify;\">Putting this in equation (2), we get<\/p>\n<p style=\"text-align: justify;\">2 (5 \u2212 <em>y<\/em>) \u2212 3<em>y <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 10 \u2212 2<em>y <\/em>\u2212 3<em>y <\/em>= 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 5<em>y <\/em>= 6 \u21d2 <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting value of y in (1), we get<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>= 5 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image003.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image003.png\" \/> and <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image002.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> 3x + 4y = 10\u2026 (1)<\/p>\n<p style=\"text-align: justify;\">2x \u2013 2y = 2\u2026 (2)<\/p>\n<p style=\"text-align: justify;\"><strong>Elimination method:<\/strong><\/p>\n<p style=\"text-align: justify;\">Multiplying equation (2) by 2, we get (3)<\/p>\n<p style=\"text-align: justify;\">4<em>x <\/em>\u2212 4<em>y <\/em>= 4 \u2026 (3)<\/p>\n<p style=\"text-align: justify;\">3<em>x <\/em>+ 4<em>y <\/em>= 10 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Adding (3) and (1), we get<\/p>\n<p style=\"text-align: justify;\">7<em>x <\/em>= 14\u21d2 <em>x <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Putting value of x in (1), we get<\/p>\n<p style=\"text-align: justify;\">3 (2) + 4<em>y <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 4<em>y <\/em>= 10 \u2013 6 = 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>y <\/em>= 1<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= 2 and <em>y <\/em>= 1<\/p>\n<p style=\"text-align: justify;\"><strong>Substitution method:<\/strong><\/p>\n<p style=\"text-align: justify;\">3<em>x <\/em>+ 4<em>y <\/em>= 10\u2026 (1)<\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>\u2212 2<em>y <\/em>= 2\u2026 (2)<\/p>\n<p style=\"text-align: justify;\">From equation (2), we get<\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>= 2 + 2<em>y<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= 1 + <em>y \u2026 <\/em>(3)<\/p>\n<p style=\"text-align: justify;\">Putting this in equation (1), we get<\/p>\n<p style=\"text-align: justify;\">3 (1 + <em>y<\/em>) + 4<em>y <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 3 + 3<em>y <\/em>+ 4<em>y <\/em>= 10<\/p>\n<p style=\"text-align: justify;\">\u21d2 7<em>y <\/em>= 7\u21d2 <em>y <\/em>= 1<\/p>\n<p style=\"text-align: justify;\">Putting value of y in (3), we get <em>x <\/em>= 1 + 1 = 2<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= 2 and <em>y <\/em>= 1<\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> 3<em>x <\/em>\u2212 5<em>y <\/em>\u2013 4 = 0 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">9<em>x <\/em>= 2<em>y <\/em>+ 7\u2026 (2)<\/p>\n<p style=\"text-align: justify;\"><strong>Elimination method:<\/strong><\/p>\n<p style=\"text-align: justify;\">Multiplying (1) by 3, we get (3)<\/p>\n<p style=\"text-align: justify;\">9<em>x <\/em>\u2212 15<em>y <\/em>\u2013 12 = 0\u2026 (3)<\/p>\n<p style=\"text-align: justify;\">9<em>x <\/em>\u2212 2<em>y <\/em>\u2013 7 = 0\u2026 (2)<\/p>\n<p style=\"text-align: justify;\">Subtracting (2) from (3), we get<\/p>\n<p style=\"text-align: justify;\">\u221213<em>y <\/em>\u2013 5 = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 \u221213<em>y <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting value of y in (1), we get<\/p>\n<p style=\"text-align: justify;\">3<em>x <\/em>\u2013 5 <img decoding=\"async\" style=\"height: 45px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image005.png\" \/>\u2212 4 = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 3<em>x <\/em>= 4 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image006.png\" \/>= <img decoding=\"async\" style=\"height: 42px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image009.png\" \/> and <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image004.png\" \/><\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Substitution_Method\"><\/span><strong>Substitution Method:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">3<em>x <\/em>\u2212 5<em>y <\/em>\u2013 4 = 0 \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">9<em>x <\/em>= 2<em>y <\/em>+ 7\u2026 (2)<\/p>\n<p style=\"text-align: justify;\">From equation (1), we can say that<\/p>\n<p style=\"text-align: justify;\">3<em>x <\/em>= 4 + 5<em>y<\/em>\u21d2 <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting this in equation (2), we get<\/p>\n<p style=\"text-align: justify;\">9 <img decoding=\"async\" style=\"height: 45px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image011.png\" \/>\u2212 2<em>y <\/em>= 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 12 + 15<em>y <\/em>\u2212 2<em>y <\/em>= 7<\/p>\n<p style=\"text-align: justify;\">\u21d2 13<em>y <\/em>= \u22125 \u21d2 <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting value of y in (1), we get<\/p>\n<p style=\"text-align: justify;\">3<em>x <\/em>\u2013 5 <img decoding=\"async\" style=\"height: 45px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image005.png\" \/>= 4<\/p>\n<p style=\"text-align: justify;\">\u21d2 3<em>x <\/em>= 4 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image006.png\" \/>= <img decoding=\"async\" style=\"height: 42px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image009.png\" \/> and <em>y <\/em>= <img decoding=\"async\" style=\"height: 42px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<\/strong> <img decoding=\"async\" style=\"height: 42px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image012.png\" \/>\u2026 (1)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image013.png\" \/>\u2026 (2)<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Elimination_method\"><\/span><strong>Elimination method:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">Multiplying equation (2) by 2, we get (3)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 80px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image014.png\" \/>\u2026 (3)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image012.png\" \/>\u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Adding (3) and (1), we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image015.png\" \/> \u21d2 <em>x <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Putting value of x in (2), we get<\/p>\n<p style=\"text-align: justify;\">2 \u2212 <img decoding=\"async\" style=\"height: 42px; width: 18px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image016.png\" \/>= 3<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>y <\/em>= \u22123<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= 2 and <em>y <\/em>= \u22123<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Substitution_method-2\"><\/span><strong>Substitution method:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image012.png\" \/>\u2026 (1)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image013.png\" \/>\u2026 (2)<\/p>\n<p style=\"text-align: justify;\">From equation (2), we can say that <img decoding=\"async\" style=\"height: 42px; width: 112px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image017.png\" \/><\/p>\n<p style=\"text-align: justify;\">Putting this in equation (1), we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image019.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 5<em>y <\/em>+ 9 = \u22126<\/p>\n<p style=\"text-align: justify;\">\u21d2 5<em>y <\/em>= \u221215\u21d2 <em>y <\/em>= \u22123<\/p>\n<p style=\"text-align: justify;\">Putting value of y in (1), we get<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 42px; width: 102px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image020.png\" \/><\/em>\u21d2 <em>x <\/em>= 2<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>x <\/em>= 2 and <em>y <\/em>= \u22123<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Form_the_pair_of_linear_equations_in_the_following_problems_and_find_their_solutions_if_they_exist_by_the_elimination_method\"><\/span><strong>2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \u00bd if we only add 1 to the denominator. What is the fraction?<\/strong><\/p>\n<p style=\"text-align: justify;\"><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><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) The sum of the digits of a two\u2013digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier togive 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><\/p>\n<p style=\"text-align: justify;\"><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 the book she kept for five days. Find the fixed charge and the charge for each extra day.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> Let numerator =<em>x <\/em>and let denominator =<em>y<\/em><\/p>\n<p style=\"text-align: justify;\">According to given condition, we have<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 43px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image021.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>+ 1 = <em>y <\/em>\u2013 1 and 2<em>x <\/em>= <em>y <\/em>+ 1<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>\u2013 <em>y <\/em>= \u22122 \u2026 <strong>(1)<\/strong> and 2<em>x <\/em>\u2013 <em>y <\/em>= 1\u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">So, we have equations <strong>(1)<\/strong> and <strong>(2)<\/strong>, multiplying equation <strong>(1)<\/strong> by 2 we get <strong>(3)<\/strong><\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>\u2212 2<em>y <\/em>= \u22124\u2026 <strong>(3)<\/strong><\/p>\n<p style=\"text-align: justify;\">2<em>x <\/em>\u2013 <em>y <\/em>= 1\u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">Subtracting equation <strong>(2)<\/strong> from <strong>(3)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">\u2212<em>y <\/em>= \u22125\u21d2 <em>y <\/em>= 5<\/p>\n<p style=\"text-align: justify;\">Putting value of y in <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>\u2013 5 = \u22122\u21d2 <em>x <\/em>= \u22122 + 5 = 3<\/p>\n<p style=\"text-align: justify;\">Therefore, fraction = <img decoding=\"async\" style=\"height: 43px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch03\/Ex3.4\/image022.png\" \/><\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<p style=\"text-align: justify;\"><strong>(ii) <\/strong>Let present age of Nuri = <em>x<\/em> years and let present age of Sonu = <em>y<\/em> years<\/p>\n<p style=\"text-align: justify;\">5 years ago, age of Nuri = (x \u2013 5) years<\/p>\n<p style=\"text-align: justify;\">5 years ago, age of Sonu = (y \u2013 5) years<\/p>\n<p style=\"text-align: justify;\">According to given condition, we have<\/p>\n<p style=\"text-align: justify;\">(<em>x <\/em>\u2212 5) = 3 (<em>y <\/em>\u2212 5)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>\u2013 5 = 3<em>y <\/em>\u2013 15<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>\u2212 3<em>y <\/em>= \u221210\u2026 <strong>(1)<\/strong><\/p>\n<p style=\"text-align: justify;\">10 years later from present, age of Nuri = (<em>x <\/em>+ 10) years<\/p>\n<p style=\"text-align: justify;\">10 years later from present, age of Sonu = (<em>y <\/em>+ 10) years<\/p>\n<p style=\"text-align: justify;\">According to given condition, we have<\/p>\n<p style=\"text-align: justify;\">(<em>x <\/em>+ 10) = 2 (<em>y <\/em>+ 10)<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>+ 10 = 2<em>y <\/em>+ 20<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>\u2212 2<em>y <\/em>= 10 \u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">Subtracting equation <strong>(1)<\/strong> from <strong>(2)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\"><em>y <\/em>= 10 \u2212 (\u221210) = 20 years<\/p>\n<p style=\"text-align: justify;\">Putting value of <strong>y<\/strong> in <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>\u2013 3 (20) = \u221210<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>\u2013 60 = \u221210<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= 50 years<\/p>\n<p style=\"text-align: justify;\">Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<p style=\"text-align: justify;\"><strong>(iii) <\/strong>Let digit at ten\u2019s place = <em>x <\/em>and Let digit at one\u2019s place = <em>y<\/em><\/p>\n<p style=\"text-align: justify;\">According to given condition, we have<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ <em>y <\/em>= 9 \u2026 <strong>(1)<\/strong><\/p>\n<p style=\"text-align: justify;\">And 9 (10<em>x <\/em>+ <em>y<\/em>) = 2 (10<em>y <\/em>+ <em>x<\/em>)<\/p>\n<p style=\"text-align: justify;\">\u21d2 90<em>x <\/em>+ 9<em>y <\/em>= 20<em>y <\/em>+ 2<em>x<\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 88<em>x <\/em>= 11<em>y <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 8<em>x <\/em>= <em>y <\/em><\/p>\n<p style=\"text-align: justify;\">\u21d2 8<em>x <\/em>\u2013 <em>y <\/em>= 0 \u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">Adding <strong>(1)<\/strong> and <strong>(2)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">9<em>x <\/em>= 9\u21d2 <em>x <\/em>= 1<\/p>\n<p style=\"text-align: justify;\">Putting value of <strong>x<\/strong> in <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">1 + <em>y <\/em>= 9<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>y <\/em>= 9 \u2013 1 = 8<\/p>\n<p style=\"text-align: justify;\">Therefore, number = 10<em>x <\/em>+ <em>y <\/em>= 10 (1) + 8 = 10 + 8 = 18<\/p>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 3.4<\/p>\n<p style=\"text-align: justify;\"><strong>(iv) <\/strong>Let number of Rs 100 notes = <em>x <\/em>and let number of Rs 50 notes = <em>y<\/em><\/p>\n<p style=\"text-align: justify;\">According to given conditions, we have<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ <em>y <\/em>= 25 \u2026 <strong>(1)<\/strong><\/p>\n<p style=\"text-align: justify;\">and 100<em>x <\/em>+ 50<em>y <\/em>= 2000<\/p>\n<p style=\"text-align: justify;\">\u21d2 2<em>x <\/em>+ <em>y <\/em>= 40 \u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">Subtracting <strong>(2)<\/strong> from <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">\u2212<em>x <\/em>= \u221215\u21d2 <em>x <\/em>= 15<\/p>\n<p style=\"text-align: justify;\">Putting value of <strong>x<\/strong> in <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">15 + <em>y <\/em>= 25<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>y <\/em>= 25 \u2013 15 = 10<\/p>\n<p style=\"text-align: justify;\">Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10<\/p>\n<p style=\"text-align: justify;\"><strong>(v) <\/strong>Let fixed charge for 3 days = Rs <em>x<\/em><\/p>\n<p style=\"text-align: justify;\">Let additional charge for each day thereafter = Rs <em>y<\/em><\/p>\n<p style=\"text-align: justify;\">According to given condition, we have<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ 4<em>y <\/em>= 27 \u2026 <strong>(1)<\/strong><\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ 2<em>y <\/em>= 21 \u2026 <strong>(2)<\/strong><\/p>\n<p style=\"text-align: justify;\">Subtracting <strong>(2)<\/strong> from <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\">2<em>y <\/em>= 6\u21d2 <em>y <\/em>= 3<\/p>\n<p style=\"text-align: justify;\">Putting value of <strong>y<\/strong> in <strong>(1)<\/strong>, we get<\/p>\n<p style=\"text-align: justify;\"><em>x <\/em>+ 4 (3) = 27<\/p>\n<p style=\"text-align: justify;\">\u21d2 <em>x <\/em>= 27 \u2013 12 = 15<\/p>\n<p style=\"text-align: justify;\">Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_34\"><\/span>NCERT Solutions for Class 10 Maths Exercise 3.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h5 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"_Download_CBSE_Question_Paper_2018\"><\/span><b><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-for-class-10-12\/\">\u00a0Download CBSE Question Paper 2018<\/a><\/strong><\/b><span class=\"ez-toc-section-end\"><\/span><\/h5>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 3.4 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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