{"id":126414,"date":"2023-09-11T08:28:00","date_gmt":"2023-09-11T02:58:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126414"},"modified":"2023-11-22T10:21:35","modified_gmt":"2023-11-22T04:51:35","slug":"rd-sharma-class-9-solutions-chapter-13-exercise-13-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127227\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\"><strong>RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/strong>: You can easily download the free PDF for <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 13<\/a> Exercise 13.2. The expert&#8217;s team has created them to help you in scoring good marks in exams.\u00a0<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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\">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=\"ez-toc-toggle-icon-1\"><label for=\"item-69e8466d06552\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" 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: #000000;color:#000000\" 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><\/label><input  type=\"checkbox\" id=\"item-69e8466d06552\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-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:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#download-rd-sharma-class-9-solutions-chapter-13-exercise-132\" title=\"Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2\">Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#access-rd-sharma-class-9-solutions-chapter-13-exercise-132\" title=\"Access RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2\">Access RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#faq-rd-sharma-class-9-solutions-chapter-13-exercise-132\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2\">FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#can-i-download-rd-sharma-class-9-solutions-chapter-13-exercise-132-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 PDF free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions?\">What are the benefits of studying RD Sharma Class 9 Solutions?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-13-exercise-132\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-13-Ex-13.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-13-Ex-13.2.pdf\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-13-exercise-132\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Write two solutions for each of the following equations:<\/strong><\/p>\n<p><strong>(i) 3x + 4y = 7<\/strong><\/p>\n<p><strong>(ii) x = 6y<\/strong><\/p>\n<p><strong>(iii) x + \u03c0y = 4<\/strong><\/p>\n<p><strong>(iv) 2\/3x \u2013 y = 4.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a03x + 4y =7 \u2026.(1)<\/p>\n<p>Step 1: Isolate the above equation in y.<\/p>\n<p>Subtract 3x from both sides,<\/p>\n<p>3x + 4y \u2013 3x = 7 \u2013 3x<\/p>\n<p>4y = 7 \u2013 3x<\/p>\n<p>Divide each side by 4<\/p>\n<p>y = 1\/4 x (7 \u2013 3x) \u2026.(2)<\/p>\n<p>Step 2: Find Solutions<\/p>\n<p>Substituting x = 1 in (2)<\/p>\n<p>y = 1\/4 x (7 \u2013 3) = 1\/4 x 4 = 1<\/p>\n<p>Thus x = 1 and y = 1 is the solution of 3x + 4y = 7<\/p>\n<p>Again, Substituting x = 2 in (2)<\/p>\n<p>y = 1\/4 x (7 \u2013 3 x 2) = 1\/4 x 1 = 1\/4<\/p>\n<p>Thus x = 2 and y = 1\/4 is the solution of 3x + 4y = 7<\/p>\n<p>Therefore, (1, 1) and (2, 1\/4) are two solutions of 3x + 4y = 7.<\/p>\n<p><strong>(ii)<\/strong>\u00a0Given: x = 6y<\/p>\n<p>Substituting x =0 in the given equation,<\/p>\n<p>0 = 6y<\/p>\n<p>or y = 0<\/p>\n<p>Thus (0,0) is one solution<\/p>\n<p>Again, substituting x=6<\/p>\n<p>6 = 6y<\/p>\n<p>or y = 1<\/p>\n<p>Thus, (6, 1) is another solution.<\/p>\n<p>Therefore, (0, 0) and (6, 1) are two solutions of x = 6y.<\/p>\n<p><strong>(iii)<\/strong>\u00a0Given: x + \u03c0y = 4<\/p>\n<p>Substituting x = 0 \u21d2 0 + \u03c0y = 4 \u21d2 y = 4\/ \u03c0<\/p>\n<p>Substituting y = 0 \u21d2 x + 0 = 4 \u21d2 x = 4<\/p>\n<p>Therefore, (0, 4\/ \u03c0) and (4, 0) are two solutions of x + \u03c0y = 4.<\/p>\n<p><strong>(iv)<\/strong>\u00a0Given: 2\/3 x \u2013 y = 4<\/p>\n<p>Substituting x = 0 \u21d2 0 \u2013 y = 4 \u21d2 y = -4<\/p>\n<p>Substituting x = 3 \u21d2 2\/3 \u00d7 3 \u2013 y = 4 \u21d2 2 \u2013 y = 4 \u21d2 y = -2<\/p>\n<p>Therefore, (0, -4) and (3, -2) are two solutions of 2\/3 x \u2013 y = 4.<\/p>\n<p><strong>Question 2: Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations :<\/strong><\/p>\n<p><strong>(i) 5x \u2013 2y =10<\/strong><\/p>\n<p><strong>(ii) -4x + 3y =12<\/strong><\/p>\n<p><strong>(iii) 2x + 3y = 24<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0Given: 5x \u2013 2y = 10<\/p>\n<p>Substituting x = 0 \u21d2 5 \u00d7 0 \u2013 2y = 10 \u21d2 \u2013 2y = 10 \u21d2 \u2013 y = 10\/2 \u21d2 y = \u2013 5<\/p>\n<p>Thus x =0 and y = -5 is the solution of 5x-2y = 10<\/p>\n<p>Substituting y = 0 \u21d2 5x \u2013 2 x 0 = 10 \u21d2 5x = 10 \u21d2 x = 2<\/p>\n<p>Thus x =2 and y = 0 is a solution of 5x \u2013 2y = 10<\/p>\n<p><strong>(ii)<\/strong>\u00a0Given, \u2013 4x + 3y = 12<\/p>\n<p>Substituting x = 0 \u21d2 -4 \u00d7 0 + 3y = 12 \u21d2 3y = 12 \u21d2 y = 4<\/p>\n<p>Thus x = 0 and y = 4 is a solution of the -4x + 3y = 12<\/p>\n<p>Substituting y = 0 \u21d2 -4 x + 3 x 0 = 12 \u21d2 \u2013 4x = 12 \u21d2 x = -3<\/p>\n<p>Thus x = -3 and y = 0 is a solution of -4x + 3y = 12<\/p>\n<p><strong>(iii)<\/strong>\u00a0Given, 2x + 3y = 24<\/p>\n<p>Substituting x = 0 \u21d2 2 x 0 + 3y = 24 \u21d2 3y =24 \u21d2 y = 8<\/p>\n<p>Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24<\/p>\n<p>Substituting y = 0 \u21d2 2x + 3 x 0 = 24 \u21d2 2x = 24 \u21d2 x =12<\/p>\n<p>Thus x = 12 and y = 0 is a solution of 2x + 3y = 24<\/p>\n<p><strong>Question 3: Check which of the following are solutions of the equation 2x \u2013 y = 6 and which are not:<\/strong><\/p>\n<p><strong>(i) ( 3 , 0 ) (ii) ( 0 , 6 ) (iii) ( 2 , -2 ) (iv) (\u221a3, 0) (v) (1\/2 , -5 )<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0Check for (3, 0)<\/p>\n<p>Put x = 3 and y = 0 in equation 2x \u2013 y = 6<\/p>\n<p>2(3) \u2013 (0) = 6<\/p>\n<p>6 = 6<\/p>\n<p>True statement.<\/p>\n<p>\u21d2 (3,0) is a solution of 2x \u2013 y = 6.<\/p>\n<p><strong>(ii)<\/strong>\u00a0Check for (0, 6)<\/p>\n<p>Put x = 0 and y = 6 in 2x \u2013 y = 6<\/p>\n<p>2 x 0 \u2013 6 = 6<\/p>\n<p>-6 = 6<\/p>\n<p>False statement.<\/p>\n<p>\u21d2 (0, 6) is not a solution of 2x \u2013 y = 6.<\/p>\n<p><strong>(iii)<\/strong>\u00a0Check for (2, -2)<\/p>\n<p>Put x = 0 and y = 6 in 2x \u2013 y = 6<\/p>\n<p>2 x 2 \u2013 (-2) = 6<\/p>\n<p>4 + 2 = 6<\/p>\n<p>6 = 6<\/p>\n<p>True statement.<\/p>\n<p>\u21d2 (2,-2) is a solution of 2x \u2013 y = 6.<\/p>\n<p><strong>(iv)<\/strong>\u00a0Check for (\u221a3, 0)<\/p>\n<p>Put x = \u221a3 and y = 0 in 2x \u2013 y = 6<\/p>\n<p>2 x \u221a3 \u2013 0 = 6<\/p>\n<p>2 \u221a3 = 6<\/p>\n<p>False statement.<\/p>\n<p>\u21d2(\u221a3, 0) is not a solution of 2x \u2013 y = 6.<\/p>\n<p><strong>(v)<\/strong>\u00a0Check for (1\/2, -5)<\/p>\n<p>Put x = 1\/2 and y = -5 in 2x \u2013 y = 6<\/p>\n<p>2 x (1\/2) \u2013 (-5) = 6<\/p>\n<p>1 + 5 = 6<\/p>\n<p>6 = 6<\/p>\n<p>True statement.<\/p>\n<p>\u21d2 (1\/2, -5) is a solution of 2x \u2013 y = 6.<\/p>\n<p><strong>Question 4: If x = -1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, 3 x + 4 y = k<\/p>\n<p>(-1, 2) is the solution of 3x + 4y = k, so it satisfies the equation.<\/p>\n<p>Substituting x = -1 and y = 2 in 3x + 4y = k, we get<\/p>\n<p>3 (\u2013 1 ) + 4( 2 ) = k<\/p>\n<p>\u2013 3 + 8 = k<\/p>\n<p>k = 5<\/p>\n<p>The value of k is 5.<\/p>\n<p><strong>Question 5: Find the value of \u03bb, if x = \u2013\u03bb and y = 5\/2 is a solution of the equation x + 4y \u2013 7 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, (-\u03bb, 5\/2) is a solution of equation 3x + 4y = k<\/p>\n<p>Substituting x = \u2013 \u03bb and y = 5\/2 in x + 4y \u2013 7 = 0, we get<\/p>\n<p>\u2013 \u03bb + 4 (5\/2) \u2013 7 =0<\/p>\n<p>-\u03bb + 10 \u2013 7 = 0<\/p>\n<p>\u03bb = 3<\/p>\n<p><strong>Question 6: If x = 2<\/strong>\u00a0<strong>\u03b1 + 1 and y = \u03b1 -1 is a solution of the equation 2x \u2013 3y + 5 = 0, find the value of \u03b1.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, (2 \u03b1 + 1, \u03b1 \u2013 1 ) is the solution of equation 2x \u2013 3y + 5 = 0.<\/p>\n<p>Substituting x = 2 \u03b1 + 1 and y = \u03b1 \u2013 1 in 2x \u2013 3y + 5 = 0, we get<\/p>\n<p>2(2 \u03b1 + 1) \u2013 3(\u03b1 \u2013 1 ) + 5 = 0<\/p>\n<p>4 \u03b1 + 2 \u2013 3 \u03b1 + 3 + 5 = 0<\/p>\n<p>\u03b1 + 10 = 0<\/p>\n<p>\u03b1 = \u2013 10<\/p>\n<p>The value of \u03b1 is -10.<\/p>\n<p><strong>Question 7: If x = 1 and y = 6 is a solution of the equation 8x \u2013 ay + a<sup>2<\/sup>\u00a0= 0, find the values of a.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, (1 , 6) is a solution of equation 8x \u2013 ay + a<sup>2<\/sup>\u00a0= 0<\/p>\n<p>Substituting x = 1 and y = 6 in 8x \u2013 ay + a<sup>2<\/sup>\u00a0= 0, we get<\/p>\n<p>8 x 1 \u2013 a x 6 + a<sup>2<\/sup>\u00a0= 0<\/p>\n<p>\u21d2 a<sup>2<\/sup>\u00a0\u2013 6a + 8 = 0 (quadratic equation)<\/p>\n<p>Using quadratic factorization<\/p>\n<p>a<sup>2<\/sup>\u00a0\u2013 4a \u2013 2a + 8 = 0<\/p>\n<p>a(a \u2013 4) \u2013 2 (a \u2013 4) = 0<\/p>\n<p>(a \u2013 2) (a \u2013 4)= 0<\/p>\n<p>a = 2, 4<\/p>\n<p>Values of a are 2 and 4.<\/p>\n<p>We have included all the information regarding\u00a0<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">. If you have any queries feel free to ask in the comment section.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-13-exercise-132\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631259302172\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-13-exercise-132-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631259433672\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631259434804\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2: You can easily download the free PDF for RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2. The expert&#8217;s team has created them to help you in scoring good marks in exams.\u00a0 RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 13 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127227,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76843],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126414"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126414"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126414\/revisions"}],"predecessor-version":[{"id":510619,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126414\/revisions\/510619"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127227"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126414"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126414"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126414"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}