{"id":126417,"date":"2023-03-31T15:01:00","date_gmt":"2023-03-31T09:31:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126417"},"modified":"2023-12-08T12:19:22","modified_gmt":"2023-12-08T06:49:22","slug":"rd-sharma-class-9-solutions-chapter-13-exercise-13-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-3\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127230\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.3.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<\/strong> are available here. 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. You can download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.1&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;}\"><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><\/span>&nbsp;Exercise 13.3 from the link provided below in this article.&nbsp;<\/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-69dbeaf623801\" 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-69dbeaf623801\"><\/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-13-3\/#download-rd-sharma-class-9-solutions-chapter-13-exercise-133\" title=\"Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3\">Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<\/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-13-3\/#access-rd-sharma-chapter-13-exercise-133-class-9-solutions\" title=\"Access RD Sharma Chapter 13 Exercise 13.3 Class 9 Solutions\">Access RD Sharma Chapter 13 Exercise 13.3 Class 9 Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-3\/#faq-rd-sharma-class-9-solutions-chapter-13-exercise-133\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3\">FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<\/a><ul class='ez-toc-list-level-3'><li class='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-13-3\/#can-i-download-rd-sharma-class-9-solutions-chapter-13-exercise-133-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3 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-13-3\/#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-13-3\/#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-133\"><\/span>Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-13.3.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-13.3.pdf\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-chapter-13-exercise-133-class-9-solutions\"><\/span>Access RD Sharma Chapter 13 Exercise 13.3 Class 9 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br>In a parallelogram ABCD, determine the sum of angles ZC and ZD.<br>Solution:<br>In a ||gm ABCD,<br>\u2220C + \u2220D = 180\u00b0<br>(Sum of consecutive angles)<br><img src=\"https:\/\/farm2.staticflickr.com\/1973\/31772237468_92bb8a5799_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"175\" height=\"156\"><\/p>\n<p>Question 2.<br>In a parallelogram ABCD, if \u2220B = 135\u00b0, determine the measures of its other angles.<br>Solution:<br>In a ||gm ABCD, \u2220B = 135\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1908\/31772237278_aeaee77da5_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"205\" height=\"152\"><br>\u2234 \u2220D = \u2220B = 135\u00b0 (Opposite angles of a ||gm)<br>But \u2220A + \u2220B = 180\u00b0 (Sum of consecutive angles)<br>\u21d2 \u2220B + 135\u00b0 = 180\u00b0<br>\u2234 \u2220A = 180\u00b0 \u2013 135\u00b0 = 45\u00b0<br>But\u2220C = \u2220B = 45\u00b0 (Opposite angles of a ||gm)<br>\u2234 Angles are 45\u00b0, 135\u00b0, 45\u00b0, 135\u00b0.<\/p>\n<p>Question 3.<br>ABCD is a square, AC and BD intersect at O. State the measure of \u2220AOB.<br>Solution:<\/p>\n<p>In a square ABCD,<br>Diagonal AC and BD intersect each other at O<br>\u2235 Diagonals of a square bisect each other at a right angle<br>\u2235\u2220AOB = 90\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1937\/31772237078_af7242d371_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"154\" height=\"161\"><\/p>\n<p>Question 4.<br>ABCD is a rectangle with \u2220ABD = 40\u00b0. Determine \u2220DBC.<br>Solution:<br>In rectangle ABCD,<br><img src=\"https:\/\/farm2.staticflickr.com\/1902\/31772236878_62efb7e80b_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"185\" height=\"155\"><br>\u2220B = 90\u00b0, BD is its diagonal<br>But \u2220ABD = 40\u00b0<br>and \u2220ABD + \u2220DBC = 90\u00b0<br>\u21d2 40\u00b0 + \u2220DBC = 90\u00b0<br>\u21d2 \u2220DBC = 90\u00b0 \u2013 40\u00b0 = 50\u00b0<br>Hence \u2220DBC = 50\u00b0<\/p>\n<p>Question 5.<br>The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.<\/p>\n<p>Solution:<br>Given: In ||gm ABCD, E, and F are the midpoints of the side AB and CD respectively<br>DE and BF are joined<br>To prove: EBFD is a ||gm<br>Construction: Join EF<br><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44730426115_c07826bc97_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"187\" height=\"152\"><br>Proof: \u2235 ABCD is a ||gm<br>\u2234 AB = CD and AB || CD<br>(Opposite sides of a ||gm are equal and parallel)<br>\u2234 EB || DF and EB = DF (\u2235 E and F are midpoints of AB and CD)<br>\u2234 EBFD is a ||gm.<\/p>\n<p>Question 6.<br>P and Q are the points of trisection of the diagonal BD of the parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.<br>Solution:<\/p>\n<p>Given: In ||gm, ABCD. P and Q are the points of trisection of the diagonal BD<br><img src=\"https:\/\/farm2.staticflickr.com\/1913\/44730425925_0bef5152ea_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"188\" height=\"153\"><br>To prove : (i) CQ || AP<br>AC bisects PQ<\/p>\n<p>Proof: \u2235 Diagonals of a parallelogram bisect each other<br>\u2234 AO = OC and BO = OD<br>\u2234 P and Q are points of trisection of BD<br>\u2234 BP = PQ = QD \u2026(i)<br>\u2235 BO = OD and BP = QD \u2026(ii)<br>Subtracting, (ii) from (i) we get<br>OB \u2013 BP = OD \u2013 QD<br>\u21d2 OP = OQ<br>In quadrilateral APCQ,<br>OA = OC and OP = OQ (proved)<br>Diagonals AC and PQ bisect each other at O<br>\u2234 APCQ is a parallelogram<br>Hence AP || CQ.<\/p>\n<p>Question 7.<br>ABCD is a square. E, F, G, and H are points on AB, BC, CD, and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.<br>Solution:<\/p>\n<p>Given: In square ABCD<br>E, F, G, and H are the points on AB, BC, CD, and DA respectively such that AE = BF = CG = DH<br>To prove: EFGH is a square<br>Proof: E, F, G, and H are points on the sides AB, BC, CA, and DA respectively such that<br>AE = BF = CG = DH = x (suppose)<br>Then BE = CF = DG = AH = y (suppose)<br>Now in \u2206AEH and \u2206BFE<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44730425695_faf719fc75_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"182\" height=\"162\"><br>AE = BF (given)<br>\u2220A = \u2220B (each 90\u00b0)<br>AH = BE (proved)<br>\u2234 \u2206AEH \u2245 \u2206BFE (SAS criterion)<br>\u2234 \u22201 = \u22202 and \u22203 = \u22204 (c.p.c.t.)<br>But \u22201 + \u22203 = 90\u00b0 and \u22202 + \u22204 = 90\u00b0 (\u2220A = \u2220B = 90\u00b0)<br>\u21d2 \u22201 + \u22202 + \u22203 + \u22204 = 90\u00b0 + 90\u00b0 = 180\u00b0<br>\u21d2 \u22201 + \u22204 + \u22201 + \u22204 = 180\u00b0<br>\u21d2 2(\u22201 + \u22204) = 180\u00b0<br>\u21d2 \u22201 + \u22204 =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-74\" class=\"math\"><span id=\"MathJax-Span-75\" class=\"mrow\"><span id=\"MathJax-Span-76\" class=\"mfrac\"><span id=\"MathJax-Span-77\" class=\"msubsup\"><span id=\"MathJax-Span-78\" class=\"texatom\"><span id=\"MathJax-Span-79\" class=\"mrow\"><span id=\"MathJax-Span-80\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-81\" class=\"texatom\"><span id=\"MathJax-Span-82\" class=\"mrow\"><span id=\"MathJax-Span-83\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-84\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 90\u00b0<br>\u2234 \u2220HEF = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<\/p>\n<p>Similarly, we can prove that<br>\u2220F = \u2220G = \u2220H = 90\u00b0<br>Since the sides of the quad. EFGH is equal and each angle is 90\u00b0<br>\u2234 EFGH is a square.<\/p>\n<p>Question 8.<br>ABCD is a rhombus, and EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.<br>Solution:<br>Given: ABCD is a rhombus, EABF is a straight line such that<br>EA = AB = BF<br>ED and FC are joined<br>Which meet at G to producing<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44730425425_7ed746a390_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"257\" height=\"150\"><br>To prove: \u2220EGF = 90\u00b0<br>Proof: \u2235 Diagonals of a rhombus bisect<br>each other at right angles<br>AO = OC, BO = OD<br>\u2220AOD = \u2220COD = 90\u00b0<br>and \u2220AOB = \u2220BOC = 90\u00b0<\/p>\n<p>In \u2206BDE,<br>A and O are the mid-points of BE and BD respectively.<br>\u2234 AO || ED<br>Similarly, OC || DG<br>In \u2206 CFA, B and O are the mid-points of AF and AC respectively<br>\u2234 OB || CF and OD || GC<br>Now in the quad. DOCG<br>OC || DG and OD || CG<br>\u2234 DOCG is a parallelogram.<br>\u2234 \u2220DGC = \u2220DOC (opposite angles of ||gm)<br>\u2234 \u2220DGC = 90\u00b0 (\u2235 \u2220DOC = 90\u00b0)<br>Hence proved.<\/p>\n<p>Question 9.<br>ABCD is a parallelogram, AD is produced to E so that DE = DC = AD and EC produced meets AB produced in F. Prove that BF = BC.<br>Solution:<\/p>\n<p>Given : In ||gm ABCD,<br>AB is produced to E so that<br>DE = DA and EC produced meets AB produced in F.<br>To prove: BF = BC<br>Proof: In \u2206ACE,<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1961\/44730425395_c3a3c0fe19_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"150\" height=\"168\"><br>O and D are the midpoints of sides AC and AE<br>\u2234 DO || EC and DB || FC<br>\u21d2 BD || EF<br>\u2234 AB = BF<br>But AB = DC (Opposite sides of ||gm)<br>\u2234 DC = BF<\/p>\n<p>Now in \u2206EDC and \u2206CBF,<br>DC = BF (proved)<br>\u2220EDC = \u2220CBF<br>(\u2235\u2220EDC = \u2220DAB corresponding angles)<br>\u2220DAB = \u2220CBF (corresponding angles)<br>\u2220ECD = \u2220CFB (corresponding angles)<br>\u2234 AEDC \u2245 ACBF (ASA criterion)<br>\u2234 DE = BC (c.p.c.t.)<br>\u21d2 DC = BC<br>\u21d2 AB = BC<br>\u21d2 BF = BC (\u2235AB = BF proved)<br>Hence proved.<\/p>\n<p>We have included all the information regarding&nbsp;<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<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.&nbsp;<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-13-exercise-133\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631259309874\" 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-133-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.3 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 13.3 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631259441096\" 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-1631259442106\" 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 13.3 are available here. 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. You can download RD Sharma Class 9 Solutions Chapter 13&nbsp;Exercise 13.3 from the link provided below &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-3\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127230,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76844],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126417"}],"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=126417"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126417\/revisions"}],"predecessor-version":[{"id":511301,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126417\/revisions\/511301"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127230"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126417"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126417"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126417"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}