{"id":125511,"date":"2023-09-13T07:27:00","date_gmt":"2023-09-13T01:57:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125511"},"modified":"2023-11-13T11:58:20","modified_gmt":"2023-11-13T06:28:20","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-3\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125554\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.3.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3:\u00a0<\/strong>The main concepts discussed in <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> Exercise 3.3 are the method of elimination by substitution and the method of elimination by equating the coefficients. For any clarifications on these exercise questions, students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-pair-of-linear-equations-in-two-variables\/\"><strong>RD Sharma Solutions Class 10 Maths Chapter 3<\/strong><\/a> Pair of Linear Equations In Two Variables Exercise 3.3 PDF provided below.<\/p>\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-69e75344b8d52\" 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\" 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id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-33-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 Free PDF<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: 800px;}<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-Solution-Class-10-Chapter-3-Exercise-3.3.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-3-Exercise-3.3.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-33-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solve the following system of equations:<\/strong><\/p>\n<p><strong>1.\u00a011x + 15y + 23 = 0 \u00a0<\/strong><\/p>\n<p><strong>7x \u2013 2y \u2013 20 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>11x +15y + 23 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>7x \u2013 2y \u2013 20 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From (ii),<\/p>\n<p>2y = 7x \u2013 20<\/p>\n<p>\u21d2 y =\u00a0(7x \u221220)\/2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>Now, substituting y in equation (i), we get<\/p>\n<p>\u21d2 11x + 15((7x\u221220)\/2) + 23 = 0<\/p>\n<p>\u21d2 11x + (105x \u2212 300)\/2 + 23 = 0<\/p>\n<p>\u21d2 (22x + 105x \u2013 300 + 46) = 0<\/p>\n<p>\u21d2 127x \u2013 254 = 0<\/p>\n<p>\u21d2 x = 2<\/p>\n<p>Next, putting the value of x in equation (iii), we get,<\/p>\n<p>\u21d2 y =\u00a0(7(2) \u2212 20)\/2<\/p>\n<p>\u2234 y= -3<\/p>\n<p>Thus, the value of x and y is found to be 2 and -3, respectively.<\/p>\n<p><strong>2. \u00a03x \u2013 7y + 10 = 0<\/strong><\/p>\n<p><strong>y \u2013 2x \u2013 3 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>3x \u2013 7y + 10 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>y \u2013 2x \u2013 3 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From (ii),<\/p>\n<p>y \u2013 2x \u2013 3 = 0<\/p>\n<p>y = 2x+3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>Now, substituting y in equation (i), we get<\/p>\n<p>\u21d2 3x \u2013 7(2x+3) + 10 = 0<\/p>\n<p>\u21d2 3x \u2013 14x \u2013 21 + 10 = 0<\/p>\n<p>\u21d2 -11x = 11<\/p>\n<p>\u21d2 x = -1<\/p>\n<p>Next, putting the value of x in equation (iii), we get<\/p>\n<p>\u21d2 y = 2(-1) + 3<\/p>\n<p>\u2234 y= 1<\/p>\n<p>Thus, the value of x and y is found to be -1 and 1, respectively.<\/p>\n<p><strong>3. 0.4x + 0.3y = 1.7<\/strong><\/p>\n<p><strong>0.7x \u2013 0.2y = 0.8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>0.4x + 0.3y = 1.7<\/p>\n<p>0.7x \u2013 0.2y = 0.8<\/p>\n<p>Let\u2019s, multiply LHS and RHS by 10 to make the coefficients an integer.<\/p>\n<p>4x + 3y = 17 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>7x \u2013 2y = 8 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>From (ii),<\/p>\n<p>7x \u2013 2y = 8<\/p>\n<p>x = (8 + 2y)\/7\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>Now, substituting x in equation (i), we get<\/p>\n<p>\u21d2 4[(8 + 2y)\/7] + 3y = 17<\/p>\n<p>\u21d2 32 + 8y + 21y = (17 x 7)<\/p>\n<p>\u21d2 29y = 87<\/p>\n<p>\u21d2 y = 3<\/p>\n<p>Next, putting the value of y in equation (iii), we get<\/p>\n<p>\u21d2 x =\u00a0(8 + 2(3))\/ 7<\/p>\n<p>\u21d2 x = 14\/7<\/p>\n<p>\u2234 x = 2<\/p>\n<p>Thus, the value of x and y is found to be 2 and 3, respectively.<\/p>\n<p><strong>4. x\/2 + y = 0.8<\/strong><\/p>\n<p><strong>7\/(x+ y\/2) = 10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>x\/2 + y = 0.8 \u21d2 x + 2y = 1.6\u2026\u2026 (a)<\/p>\n<p>7\/(x + y\/2) = 10 \u21d27 = 10(x + y\/2) \u21d27 = 10x + 5y<\/p>\n<p>Let\u2019s, multiply the LHS and RHS of equation (a) by 10 for easy calculation.<\/p>\n<p>So, we finally get<\/p>\n<p>10x + 20y = 16 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i) And,<\/p>\n<p>10x + 5y = 7 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>Now, subtracting two equations, we get<\/p>\n<p>\u21d2 (i) \u2013 (ii)<\/p>\n<p>15y = 9<\/p>\n<p>\u21d2 y = 3\/5<\/p>\n<p>Next, putting the value of y in the equation (i), we get<\/p>\n<p>x = [16 \u2212 20(3\/5)]\/10<\/p>\n<p>\u21d2\u00a0 (16 \u2013 12)\/10 = 4\/10<\/p>\n<p>\u2234 x =\u00a02\/5<\/p>\n<p>Thus, the value of x and y obtained are 2\/5 and 3\/5, respectively.<\/p>\n<p><strong>5. 7(y + 3) \u2013 2(x + 2) = 14<\/strong><\/p>\n<p><strong>4(y \u2013 2) + 3(x \u2013 3) = 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>7(y+3) \u2013 2(x+2) = 14\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>4(y-2) + 3(x-3) = 2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From (i), we get<\/p>\n<p>7y + 21 \u2013 2x \u2013 4 = 14<\/p>\n<p>7y = 14 + 4 \u2013 21 + 2x<\/p>\n<p>\u21d2 y = (2x \u2013 3)\/7<\/p>\n<p>From (ii), we get<\/p>\n<p>4y \u2013 8 + 3x \u2013 9 = 2<\/p>\n<p>4y + 3x \u2013 17 \u2013 2 = 0<\/p>\n<p>\u21d2 4y + 3x \u2013 19 = 0 \u2026\u2026\u2026\u2026\u2026.. (iii)<\/p>\n<p>Now, substituting y in equation (iii),<\/p>\n<p>4[(2x \u2212 3)\/7] + 3x \u2013 19=0<\/p>\n<p>8x \u2013 12 + 21x \u2013 (19 x 17) = 0 [after taking LCM]<\/p>\n<p>29x = 145<\/p>\n<p>\u21d2 x = 5<\/p>\n<p>Now, putting the value of x and in equation (ii),<\/p>\n<p>4(y-2) + 3(5-3) = 2<\/p>\n<p>\u21d2 4y -8 + 6 = 2<\/p>\n<p>\u21d2 4y = 4<\/p>\n<p>\u2234 y =\u00a01<\/p>\n<p>Thus, the value of x and y obtained are 5 and 1, respectively.<\/p>\n<p><strong>6. x\/7 + y\/3 = 5<\/strong><\/p>\n<p><strong>x\/2 \u2013 y\/9 = 6<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>x\/7 + y\/3 = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>x\/2 \u2013 y\/9 = 6\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)<\/p>\n<p>From (i), we get<\/p>\n<p>x\/7 + y\/3 = 5<\/p>\n<p>\u21d23x + 7y = (5\u00d721) [After taking LCM]<\/p>\n<p>\u21d2 3x =105 \u2013 7y<\/p>\n<p>\u21d2 x = (105 \u2013 7y)\/3\u2026\u2026. (iv)<\/p>\n<p>From (ii), we get<\/p>\n<p>x\/2 \u2013 y\/9 = 6<\/p>\n<p>\u21d2 9x \u2013 2y = 108 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii) [After taking LCM]<\/p>\n<p>Now, substituting x in equation (iii), we get<\/p>\n<p>9[(105 \u2212 7y)\/3] \u2013 2y = 108<\/p>\n<p>\u21d2 945 \u2013 63y \u2013 6y = 324 [After taking LCM]<\/p>\n<p>\u21d2 945 \u2013 324 = 69y<\/p>\n<p>\u21d2 69y = 621<\/p>\n<p>\u21d2 y = 9<\/p>\n<p>Now, putting the value of y in equation (iv),<\/p>\n<p>x = (105 \u2212 7(9))\/3<\/p>\n<p>\u21d2 x = (105 \u2212 63)\/3 = 42\/3<\/p>\n<p>\u2234 x = 14<\/p>\n<p>Thus, the value of x and y obtained are 14 and 9, respectively.<\/p>\n<p><strong>7. x\/3 + y\/4 = 11<\/strong><\/p>\n<p><strong>5x\/6 \u2212 y\/3 = \u22127<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>x\/3 + y\/4 = 11\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>5x\/6 \u2212 y\/3 = \u22127\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From (i), we get<\/p>\n<p>x\/3 + y\/4 = 11<\/p>\n<p>\u21d24x + 3y = (11\u00d712) [After taking LCM]<\/p>\n<p>\u21d2 4x =132 \u2013 3y<\/p>\n<p>\u21d2 x = (132 \u2013 3y)\/4\u2026\u2026. (iv)<\/p>\n<p>From (ii), we get<\/p>\n<p>5x\/6 \u2212 y\/3 = \u22127<\/p>\n<p>\u21d2 5x \u2013 2y = -42 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii) [After taking LCM]<\/p>\n<p>Now, substituting x in equation (iii), we get<\/p>\n<p>5[(132 \u2212 3y)\/4] \u2013 2y = -42<\/p>\n<p>\u21d2 660 \u2013 15y \u2013 8y = -42 x 4 [After taking LCM]<\/p>\n<p>\u21d2 660 + 168 = 23y<\/p>\n<p>\u21d2 23y = 828<\/p>\n<p>\u21d2 y = 36<\/p>\n<p>Now, putting the value of y in equation (iv),<\/p>\n<p>x = (132 \u2013 3(36))\/4<\/p>\n<p>\u21d2 x = (132 \u2212 108)\/4 = 24\/4<\/p>\n<p>\u2234 x = 6<\/p>\n<p>Thus, the value of x and y obtained are 6 and 36, respectively.<\/p>\n<p><strong>8. 4\/x + 3y = 8<\/strong><\/p>\n<p><strong>6\/x \u22124y = \u22125<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking 1\/x = u<\/p>\n<p>Then, the two equations become<\/p>\n<p>4u + 3y = 8\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i)<\/p>\n<p>6u \u2013 4y = -5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>From (i), we get<\/p>\n<p>4u = 8 \u2013 3y<\/p>\n<p>\u21d2 u =\u00a0(8 \u2212 3y)\/4 \u2026\u2026.. (iii)<\/p>\n<p>Substituting u in (ii),<\/p>\n<p>[6(8 \u2212 3y)\/4] \u2013 4y = -5<\/p>\n<p>\u21d2\u00a0 [3(8\u22123y)\/2] \u2212 4y = \u22125<\/p>\n<p>\u21d2 24 \u2212 9y \u22128y = \u22125 x 2 [After taking LCM]<\/p>\n<p>\u21d2 24 \u2013 17y = -10<\/p>\n<p>\u21d2 -17y =- 34<\/p>\n<p>\u21d2 y = 2<\/p>\n<p>Putting y=2 in (iii), we get<\/p>\n<p>u = (8 \u2212 3(2))\/4<\/p>\n<p>\u21d2 u = (8 \u2212 6)\/4<\/p>\n<p>\u21d2 u = 2\/4 = 1\/2<\/p>\n<p>\u21d2 x = 1\/u = 2<\/p>\n<p>\u2234 x = 2<\/p>\n<p>So, the solution of\u00a0the pair of equations given is\u00a0x=2 and y =2.<\/p>\n<p><strong>9. x + y\/2 = 4<\/strong><\/p>\n<p><strong>2y + x\/3 = 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>x + y\/2 = 4\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>2y + x\/3 = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>From (i), we get<\/p>\n<p>x + y\/2 = 4<\/p>\n<p>\u21d2 2x + y = 8 [After taking LCM]<\/p>\n<p>\u21d2 y = 8 \u2013 2x \u2026..(iv)<\/p>\n<p>From (ii), we get<\/p>\n<p>x + 6y = 15 \u2026\u2026\u2026\u2026\u2026\u2026 (iii) [After taking LCM]<\/p>\n<p>Substituting y in (iii), we get<\/p>\n<p>x + 6(8 \u2013 2x) = 15<\/p>\n<p>\u21d2 x + 48 \u2013 12x = 15<\/p>\n<p>\u21d2 -11x = 15 \u2013 48<\/p>\n<p>\u21d2 -11x = -33<\/p>\n<p>\u21d2 x = 3<\/p>\n<p>Putting x = 3 in (iv), we get<\/p>\n<p>y = 8 \u2013 (2\u00d73)<\/p>\n<p>\u2234 y = 8 \u2013 6 = 2<\/p>\n<p>Hence, the solutions of the given system of the equation are x = 3 and y = 2, respectively.<\/p>\n<p><strong>10. x + 2y =\u00a03\/2<\/strong><\/p>\n<p><strong>2x + y =\u00a03\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>x + 2y =\u00a03\/2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>2x + y =\u00a03\/2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1, respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.<\/p>\n<p>Multiplying equation (i)x1 and (ii)x2 \u21d2<\/p>\n<p>x + 2y =\u00a03\/2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iii)<\/p>\n<p>4x + 2y = 3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iv)<\/p>\n<p>Subtracting equation (iii) from (iv),<\/p>\n<p>(4x \u2013 x) + (2y-2y) = 3x =\u00a03 \u2013 (3\/2)<\/p>\n<p>\u21d2 3x =\u00a03\/2<\/p>\n<p>\u21d2 x = 1\/2<\/p>\n<p>Putting x = 1\/2 in equation (iv),<\/p>\n<p>4(1\/2) + 2y = 3<\/p>\n<p>\u21d2 2 + 2y = 3<\/p>\n<p>\u2234 y=\u00a01\/2<\/p>\n<p>The solution of the system of equations is x = 1\/2 and y = 1\/2<\/p>\n<p><strong>11. \u221a2x \u2013 \u221a3y = 0<\/strong><\/p>\n<p><strong>\u221a3x \u2212 \u221a8y = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>\u221a2x \u2013 \u221a3y = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>\u221a3x \u2212 \u221a8y = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From equation (i),<\/p>\n<p>x = \u221a(3\/2)y\u00a0\u2026\u2026\u2026\u2026\u2026..(iii)<\/p>\n<p>Substituting this value in equation (ii), we obtain<\/p>\n<p>\u221a3x \u2212 \u221a8y = 0<\/p>\n<p>\u21d2 \u221a3(\u221a(3\/2)y)\u00a0 \u2212 \u221a8y = 0<\/p>\n<p>\u21d2 (3\/\u221a2)y \u2013 \u221a8y = 0<\/p>\n<p>\u21d2 3y \u2013 4y = 0<\/p>\n<p>\u21d2 y = 0<\/p>\n<p>Now, substituting y in equation (iii), we obtain<\/p>\n<p>\u21d2 x=0<\/p>\n<p>Thus, the value of x and y obtained are 0 and 0, respectively.<\/p>\n<p><strong>12. 3x \u2013 (y + 7)\/11 + 2 = 10<\/strong><\/p>\n<p><strong>2y + (x + 11)\/7 = 10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>3x \u2013 (y + 7)\/11 + 2 = 10\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>2y + (x + 11)\/7 = 10\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From equation (i),<\/p>\n<p>33x \u2013 y \u2013 7 + 22 = (10 x 11) [After taking LCM]<\/p>\n<p>\u21d2 33x \u2013 y + 15 = 110<\/p>\n<p>\u21d2 33x + 15 \u2013 110 = y<\/p>\n<p>\u21d2 y = 33x \u2013 95\u2026\u2026\u2026. (iv)<\/p>\n<p>From equation (ii),<\/p>\n<p>14 + x + 11 = (10 x 7) [After taking LCM]<\/p>\n<p>\u21d2 14y + x + 11 = 70<\/p>\n<p>\u21d2 14y + x = 70 \u2013 11<\/p>\n<p>\u21d2 14y + x = 59 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)<\/p>\n<p>Substituting (iv) in (iii), we get<\/p>\n<p>14 (33x \u2013 95) + x = 59<\/p>\n<p>\u21d2 462x \u2013 1330 + x = 59<\/p>\n<p>\u21d2 463x = 1389<\/p>\n<p>\u21d2 x = 3<\/p>\n<p>Putting x = 3 in (iii), we get<\/p>\n<p>\u21d2 y = 33(3) \u2013 95<\/p>\n<p>\u2234 y= 4<\/p>\n<p>The solution for the given pair of equations is x = 3 and y = 4, respectively.<\/p>\n<p><strong>13. 2x \u2013 (3\/y) = 9<\/strong><\/p>\n<p><strong>3x + (7\/y) = 2 ,y \u2260 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>2x \u2013 (3\/y) = 9\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>3x + (7\/y) = 2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>Substituting 1\/y = u, the above equations become<\/p>\n<p>2x \u2013 3u = 9 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iii)<\/p>\n<p>3x + 7u = 2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iv)<\/p>\n<p>From (iii)<\/p>\n<p>2x = 9 + 3u<\/p>\n<p>\u21d2 x =\u00a0(9+3u)\/2<\/p>\n<p>Substituting the value of x from above in equation (iv), we get<\/p>\n<p>3[(9+3u)\/2] + 7u = 2<\/p>\n<p>\u21d2 27 + 9u + 14u = (2 x 2)<\/p>\n<p>\u21d2 27 + 23u = 4<\/p>\n<p>\u21d2 23u = -23<\/p>\n<p>\u21d2 u = -1<\/p>\n<p>So, y = 1\/u\u00a0= -1<\/p>\n<p>And putting u = -1 in x = (9 + 3u)\/2, we get<\/p>\n<p>\u21d2 x =\u00a0[9 + 3(\u22121)]\/2 = 6\/2<\/p>\n<p>\u2234 x = 3<\/p>\n<p>The solutions of the pair of equations given are y = 3 and x = -1, respectively.<\/p>\n<p><strong>14. 0.5x + 0.7y = 0.74<\/strong><\/p>\n<p><strong>0.3x + 0.5y = 0.5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>0.5x + 0.7y = 0.74\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i)<\/p>\n<p>0.3x \u2013 0.5y = 0.5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Now, let\u2019s multiply LHS and RHS by 100 for both (i) and (ii) to make integral coefficients and constants.<\/p>\n<p>(i) x100 \u21d2<\/p>\n<p>50x +70y = 74 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)<\/p>\n<p>(ii) x100 \u21d2<\/p>\n<p>30x + 50y = 50 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<\/p>\n<p>From (iii),<\/p>\n<p>50x = 74 \u2013 70y<\/p>\n<p>x = (74\u221270y)\/ 50 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (v)<\/p>\n<p>Now, substituting x in equation (iv), we get<\/p>\n<p>30[(74\u221270y)\/ 50] + 50y = 50<\/p>\n<p>\u21d2 222 \u2013 210y + 250y = 250 [After taking LCM]<\/p>\n<p>\u21d2 40y = 28<\/p>\n<p>\u21d2 y = 0.7<\/p>\n<p>Now, by putting the value of\u00a0y in the equation (v), we get<\/p>\n<p>\u21d2 x =\u00a0[74 \u2212 70(0.7)]\/ 50=0<\/p>\n<p>\u21d2 x =25\/ 50 = 1\/2<\/p>\n<p>\u2234 x = 0.5<\/p>\n<p>Thus, the values of x and y so obtained are 0.5 and 0.7, respectively.<\/p>\n<p><strong>15. 1\/(7x) + 1\/(6y) = 3<\/strong><\/p>\n<p><strong>1\/(2x) \u2013 1\/(3y) = 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>1\/(7x) + 1\/(6y) = 3\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>1\/(2x) \u2013 1\/(3y) = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Multiplying (ii) by 1\/2, we get<\/p>\n<p>1\/(4x) \u2013 1\/(6y) = 52\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iii)<\/p>\n<p>Now, solving equations (i) and (iii),<\/p>\n<p>1\/(7x) + 1\/(6y) = 3\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>1\/(4x) \u2013 1\/(6y) = 5\/2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iii)<\/p>\n<p>Adding (i) + (iii), we get<\/p>\n<p>1\/x(1\/7 + 1\/4 ) = 3 + 5\/2<\/p>\n<p>\u21d2 1\/x(11\/28) = 11\/2<\/p>\n<p>\u21d2 x = 1\/14<\/p>\n<p>Using x =1\/ 14, we get, in (i)<\/p>\n<p>1\/[7(1\/14)] + 1\/(6y) = 3<\/p>\n<p>\u21d2 2 + 1\/(6y)=3<\/p>\n<p>\u21d2 1\/(6y) = 1<\/p>\n<p>\u21d2 y = 1\/6<\/p>\n<p>The solution for the given pair of equations is x=1\/14 and y=1\/6, respectively.<\/p>\n<p><strong>16. 1\/(2x) + 1\/(3y) = 2<\/strong><\/p>\n<p><strong>\u00a0 1\/(3x) + 1\/(2y) = 13\/6<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and\u00a01\/y\u00a0= v,<\/p>\n<p>So, the given equations become<\/p>\n<p>u\/2 + v\/3 = 2 \u2026\u2026\u2026\u2026\u2026\u2026(i)<\/p>\n<p>u\/3 + v\/2 = 13\/6 \u2026\u2026\u2026\u2026\u2026(ii)<\/p>\n<p>From (i), we get<\/p>\n<p>u\/2 + v\/3 = 2<\/p>\n<p>\u21d2 3u + 2v = 12<\/p>\n<p>\u21d2 u = (12 \u2013 2v)\/3 \u2026\u2026\u2026\u2026.(iii)<\/p>\n<p>Using (iii) in (ii),<\/p>\n<p>[(12 \u2013 2v)\/3]\/3 + v\/2 = 13\/6<\/p>\n<p>\u21d2 (12 \u2013 2v)\/9 + v\/2 = 13\/6<\/p>\n<p>\u21d2 24 \u2013 4v + 9v = (13\/6) x 18 [after taking LCM]<\/p>\n<p>\u21d2 24 + 5v = 39<\/p>\n<p>\u21d2 5v = 15<\/p>\n<p>\u21d2 v = 3<\/p>\n<p>Substituting v in (iii),<\/p>\n<p>u = (12 \u2013 2(3))\/3<\/p>\n<p>\u21d2 u = 2<\/p>\n<p>Thus, x = 1\/u \u21d2 x = 1\/2 and<\/p>\n<p>y = 1\/v \u21d2 y = 1\/3<\/p>\n<p>The solution for the given pair of equations is x = 1\/2 and y = 1\/3, respectively.<\/p>\n<p><strong>17. 15\/u + 2\/v = 17<\/strong><\/p>\n<p><strong>1\/u + 1\/v = 36\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and 1\/y\u00a0= v<\/p>\n<p>So, the given equations become<\/p>\n<p>15x + 2y = 17 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>x + y = 36\/5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>From equation (i), we get<\/p>\n<p>2y = 17 \u2013 15x<\/p>\n<p>=y =\u00a0(17 \u2212 15x)\/ 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iii)<\/p>\n<p>Substituting (iii) in equation (ii), we get<\/p>\n<p>= x + (17 \u2212 15x)\/ 2 =\u00a036\/5<\/p>\n<p>2x + 17 \u2013 15x = (36 x 2)\/ 5 [after taking LCM]<\/p>\n<p>-13x = 72\/5 \u2013 17<\/p>\n<p>= -13x = -13\/5<\/p>\n<p>\u21d2 x = 1\/5<\/p>\n<p>\u21d2 u = 1\/x = 5<\/p>\n<p>Putting x = 1\/5in equation (ii), we get<\/p>\n<p>1\/5\u00a0+ y = 36\/5<\/p>\n<p>\u21d2 y = 7<\/p>\n<p>\u21d2 v = 1\/y =\u00a01\/7<\/p>\n<p>The solutions of the pair of equations given are u = 5 and v = 1\/7, respectively.<\/p>\n<p><strong>18. 3\/x \u2013 1\/y = \u22129<\/strong><\/p>\n<p><strong>2\/x + 3\/y = 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and 1\/y\u00a0= v<\/p>\n<p>So, the given equations become<\/p>\n<p>3u \u2013 v = -9\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)<\/p>\n<p>2u + 3v = 5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)<\/p>\n<p>Multiplying equation (i) x 3 and (ii) x 1, we get<\/p>\n<p>9u \u2013 3v = -27 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)<\/p>\n<p>2u + 3v = 5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<\/p>\n<p>Adding equations (iii) and (iv), we get<\/p>\n<p>9u + 2u \u2013 3v + 3v = -27 + 5<\/p>\n<p>\u21d2 11u = -22<\/p>\n<p>\u21d2 u = -2<\/p>\n<p>Now putting u =-2 in equation (iv), we get<\/p>\n<p>2(-2) + 3v = 5<\/p>\n<p>\u21d2 3v = 9<\/p>\n<p>\u21d2 v = 3<\/p>\n<p>Hence, x = 1\/u =\u00a0\u22121\/2 and,<\/p>\n<p>y = 1\/v =\u00a01\/3<\/p>\n<p><strong>19. 2\/x + 5\/y = 1<\/strong><\/p>\n<p><strong>60\/x + 40\/y = 19<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and 1\/y\u00a0= v<\/p>\n<p>So, the given equations become<\/p>\n<p>2u + 5v = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)<\/p>\n<p>60u + 40v = 19 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)<\/p>\n<p>Multiplying equation (i) x 8 and (ii) x 1, we get<\/p>\n<p>16u + 40v = 8 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)<\/p>\n<p>60u + 40v = 19 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<\/p>\n<p>Subtracting equation (iii) from (iv), we get<\/p>\n<p>60u \u2013 16u + 40v \u2013 40v = 19 \u2013 8<\/p>\n<p>\u21d2 44u = 11<\/p>\n<p>\u21d2 u = 1\/4<\/p>\n<p>Now putting u = 1\/4 in equation (iv), we get<\/p>\n<p>60(1\/4) + 40v = 19<\/p>\n<p>\u21d2 15 + 40v = 19<\/p>\n<p>\u21d2 v = 4\/ 40 = 1\/10<\/p>\n<p>Hence, x = 1\/u = 4 and<\/p>\n<p>y = 1\/v =\u00a010<\/p>\n<p><strong>20. 1\/(5x) + 1\/(6y) = 12<\/strong><\/p>\n<p><strong>1\/(3x) \u2013 3\/(7y) = 8<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and 1\/y\u00a0= v<\/p>\n<p>So, the given equations become<\/p>\n<p>u\/5 + v\/6 = 12\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)<\/p>\n<p>u\/3 \u2013 3v\/7 = 8\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)<\/p>\n<p>Taking LCM for both equations, we get<\/p>\n<p>6u + 5v = 360\u2026\u2026\u2026. (iii)<\/p>\n<p>7u \u2013 9v = 168\u2026\u2026\u2026.. (iv)<\/p>\n<p>Subtracting (iii) from (iv),<\/p>\n<p>7u \u2013 9v \u2013 (6u + 5v) = 168 \u2013 360<\/p>\n<p>\u21d2 u \u2013 14v = -192<\/p>\n<p>\u21d2 u = (14v \u2013 192)\u2026\u2026\u2026. (v)<\/p>\n<p>Using (v) in equation (iii), we get<\/p>\n<p>6(14v \u2013 192) + 5v = 360<\/p>\n<p>\u21d2 84v -1152 + 5v = 360<\/p>\n<p>\u21d2 89v = 1512<\/p>\n<p>\u21d2 v = 1512\/89<\/p>\n<p>\u21d2 y = 1\/v = 89\/1512<\/p>\n<p>Now, substituting v in equation (v), we find u.<\/p>\n<p>u = 14 x (1512\/89) \u2013 192<\/p>\n<p>\u21d2 u = 4080\/89<\/p>\n<p>\u21d2 x = 1\/u = 89\/ 4080<\/p>\n<p>Hence, the solution for the given system of equations is x = 89\/4080 and y = 89\/ 1512.<\/p>\n<p><strong>21. 4\/x + 3y = 14<\/strong><\/p>\n<p><strong>3\/x \u2013 4y = 23<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking 1\/x = u, the given equations become<\/p>\n<p>4u + 3y = 14\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>3u \u2013 4y = 23\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Adding (i) and (ii), we get<\/p>\n<p>4u + 3y + 3u \u2013 4y = 14 + 23<\/p>\n<p>\u21d2 7u \u2013 y = 37<\/p>\n<p>\u21d2 y = 7u \u2013 37\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>Using (iii) in (i),<\/p>\n<p>4u + 3(7u \u2013 37) = 14<\/p>\n<p>\u21d2 4u + 21u \u2013 111 = 14<\/p>\n<p>\u21d2 25u = 125<\/p>\n<p>\u21d2 u = 5<\/p>\n<p>\u21d2 x = 1\/u = 1\/5<\/p>\n<p>Putting u= 5 in (iii), we find y<\/p>\n<p>y = 7(5) \u2013 37<\/p>\n<p>\u21d2 y = -2<\/p>\n<p>Hence, the solution for the given system of equations is x = 1\/5 and y = -2.<\/p>\n<p><strong>22.<\/strong>\u00a0<strong>4\/x + 5y = 7<\/strong><\/p>\n<p><strong>3\/x + 4y = 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking 1\/x = u, the given equations become<\/p>\n<p>4u + 5y = 7\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>3u + 4y = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Subtracting (ii) from (i), we get<\/p>\n<p>4u + 5y \u2013 (3u + 4y) = 7 \u2013 5<\/p>\n<p>\u21d2 u + y = 2<\/p>\n<p>\u21d2 u = 2 \u2013 y\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>Using (iii) in (i),<\/p>\n<p>4(2 \u2013 y) + 5y = 7<\/p>\n<p>\u21d2 8 \u2013 4y + 5y = 7<\/p>\n<p>\u21d2 y = -1<\/p>\n<p>Putting y = -1 in (iii), we find u.<\/p>\n<p>u = 2 \u2013 (-1)<\/p>\n<p>\u21d2 u = 3<\/p>\n<p>\u21d2 x = 1\/u = 1\/3<\/p>\n<p>Hence, the solution for the given system of equations is x = 1\/3 and y = -1.<\/p>\n<p><strong>23.<\/strong>\u00a0<strong>2\/x + 3\/y = 13<\/strong><\/p>\n<p><strong>5\/x \u2013 4\/y = -2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/x\u00a0= u and 1\/y\u00a0= v<\/p>\n<p>So, the given equations become<\/p>\n<p>2u + 3v = 13\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>5u \u2013 4v = -2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Adding equations (i) and (ii), we get<\/p>\n<p>2u + 3v + 5u \u2013 4v = 13 \u2013 2<\/p>\n<p>\u21d2 7u \u2013 v = 11<\/p>\n<p>\u21d2 v = 7u \u2013 11\u2026\u2026.. (iii)<\/p>\n<p>Using (iii) in (i), we get<\/p>\n<p>2u + 3(7u \u2013 11) = 13<\/p>\n<p>\u21d2 2u + 21u \u2013 33 = 13<\/p>\n<p>\u21d2 23u = 46<\/p>\n<p>\u21d2 u = 2<\/p>\n<p>Substituting u = 2 in (iii), we find v.<\/p>\n<p>v = 7(2) \u2013 11<\/p>\n<p>\u21d2 v = 3<\/p>\n<p>Hence, x = 1\/u =\u00a01\/2 and,<\/p>\n<p>y = 1\/v =\u00a01\/3<\/p>\n<p><strong>24. 2\/x + 3\/y = 2<\/strong><\/p>\n<p><strong>4\/x \u2013 9\/y = -1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a01\/\u221ax\u00a0= u and 1\/\u221ay\u00a0= v,<\/p>\n<p>So, the given equations become<\/p>\n<p>2u + 3v = 2\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>4u \u2013 9v = -1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Multiplying (ii) by 3 and<\/p>\n<p>Adding equations (i) and (ii)x3, we get<\/p>\n<p>6u + 9v + 4u \u2013 9v = 6 \u2013 1<\/p>\n<p>\u21d2 10u = 5<\/p>\n<p>\u21d2 u = 1\/2<\/p>\n<p>Substituting u = 1\/2 in (i), we find v<\/p>\n<p>2(1\/2) + 3v = 2<\/p>\n<p>\u21d2 3v = 2 \u2013 1<\/p>\n<p>\u21d2 v = 1\/3<\/p>\n<p>Since, 1\/\u221ax = u we get x = 1\/u<sup>2<\/sup><\/p>\n<p>\u21d2 x = 1\/(1\/2)<sup>2<\/sup>\u00a0= 4<\/p>\n<p>And,<\/p>\n<p>1\/\u221ay = v y = 1\/v<sup>2<\/sup><\/p>\n<p>\u21d2 y = 1\/(1\/3)<sup>2\u00a0<\/sup>= 9<\/p>\n<p>Hence, the solution is x = 4 and y = 9.<\/p>\n<p><strong>25. (x + y)\/xy = 2<\/strong><\/p>\n<p><strong>(x \u2013 y)\/xy = 6<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The given pair of equations are<\/p>\n<p>(x + y)\/xy = 2 \u21d2 1\/y + 1\/x = 2\u2026\u2026. (i)<\/p>\n<p>(x \u2013 y)\/xy = 6 \u21d2 1\/y \u2013 1\/x = 6\u2026\u2026\u2026(ii)<\/p>\n<p>Let 1\/x = u and 1\/y = v, so the equations (i) and (ii) become<\/p>\n<p>v + u = 2\u2026\u2026. (iii)<\/p>\n<p>v \u2013 u = 6\u2026\u2026..(iv)<\/p>\n<p>Adding (iii) and (iv), we get<\/p>\n<p>2v = 8<\/p>\n<p>\u21d2 v = 4<\/p>\n<p>\u21d2 y = 1\/v = 1\/4<\/p>\n<p>Substituting v = 4 in (iii) to find x,<\/p>\n<p>4 + u = 2<\/p>\n<p>\u21d2 u = -2<\/p>\n<p>\u21d2 x = 1\/u = -1\/2<\/p>\n<p>Hence, the solution is x = -1\/2 and y = 1\/4.<\/p>\n<p><strong>26. 2\/x + 3\/y = 9\/xy<\/strong><\/p>\n<p><strong>4\/x + 9\/y = 21\/xy<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Taking LCM for both the given equations, we have<\/p>\n<p>(2y + 3x)\/ xy = 9\/xy \u21d2 3x + 2y = 9\u2026\u2026\u2026. (i)<\/p>\n<p>(4y + 9x)\/ xy = 21\/xy \u21d2 9x + 4y = 21\u2026\u2026\u2026(ii)<\/p>\n<p>Performing (ii) \u2013 (i)x2\u21d2<\/p>\n<p>9x + 4y \u2013 2(3x + 2y) = 21 \u2013 (9\u00d72)<\/p>\n<p>\u21d2 3x = 3<\/p>\n<p>\u21d2 x = 1<\/p>\n<p>Using x = 1 in (i), we find y<\/p>\n<p>3(1) + 2y = 9<\/p>\n<p>\u21d2 y = 6\/2<\/p>\n<p>\u21d2 y = 3<\/p>\n<p>Thus, the solution for the given set of equations is x = 1 and y = 3.<\/p>\n<hr \/>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-3-exercise-33\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3<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-1631107856431\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-3-exercise-33-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108074152\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-practice-all-of-the-questions-in-rd-sharma-class-10-solutions-chapter-3-exercise-33\"><\/span>Is it required to practice all of the questions in RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all of the questions in RD Sharma Solutions for Class 10 Maths Chapter 3 Exercise 3.3 must be learned. These questions may appear on both board exams and class tests. Students will be prepared for their board exams if they learn these questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108615072\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-33\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3:\u00a0The main concepts discussed in RD Sharma Class 10 Solutions Exercise 3.3 are the method of elimination by substitution and the method of elimination by equating the coefficients. For any clarifications on these exercise questions, students can use the RD Sharma Solutions Class 10 Maths Chapter 3 &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-3\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125554,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125511"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125511"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125511\/revisions"}],"predecessor-version":[{"id":506725,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125511\/revisions\/506725"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125554"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125511"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125511"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125511"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}