{"id":6019,"date":"2023-07-22T07:22:44","date_gmt":"2023-07-22T07:22:44","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6019"},"modified":"2023-07-23T05:45:40","modified_gmt":"2023-07-23T05:45:40","slug":"ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 3 (Pair of Linear Equations in Two Variables)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 3 Pair of Linear Equations in Two Variables <\/strong>Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 3 Pair of Linear Equations in Two Variables<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.4<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.5<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.6<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-7\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.7<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 3.2\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Form the pair of linear equations in the following problems, and find their solutions graphically.<br \/>\n<\/strong><strong>(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.<br \/>\n<\/strong><strong>(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i) <\/strong>Let the number of girls be <i>x <\/i>and the number of boys be <i>y<\/i>.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">According to the question, the algebraic representation is<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>x<\/i> + <i>y<\/i> = 10<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>x<\/i> \u2212 <i>y<\/i> = 4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">For <i>x<\/i> + <i>y<\/i> = 10,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>x<\/i> = 10 \u2212 <i>y<br \/>\n<\/i><\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">6<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">6<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">For <i>x<\/i> \u2212 <i>y<\/i> = 4,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><i>x<\/i> = 4 + <i>y<br \/>\n<\/i><\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">-1<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The graphical representation is as follows;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6177\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"648\" height=\"658\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1.png 648w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1-295x300.png 295w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1-300x305.png 300w\" sizes=\"auto, (max-width: 648px) 100vw, 648px\" \/><br \/>\nFrom the figure, it can be observed that these lines intersect each other at point (7, 3).<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, the number of girls and boys in the class are 7 and 3 respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>Let the cost of 1 pencil be Rs <i>x <\/i>and the cost of 1 pen be Rs <i>y<\/i>.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">According to the question, the algebraic representation is<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">5<i>x<\/i> + 7<i>y<\/i> = 50<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">7<i>x<\/i> + 5<i>y<\/i> = 46<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">For 5<i>x<\/i> + 7<i>y<\/i> = 50,<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{50-7y}{5}\" alt=\"\\frac{50-7y}{5}\" align=\"absmiddle\" \/><\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">10<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">10<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">-4<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">7<i>x<\/i> + 5<i>y<\/i> = 46<\/span><\/p>\n<p>x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{46-5y}{7}\" alt=\"\\frac{46-5y}{7}\" align=\"absmiddle\" \/><\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">-2<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">5<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">12<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">8<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">-2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, the graphic representation is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6178\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"683\" height=\"689\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1i.png 683w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1i-297x300.png 297w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1i-150x150.png 150w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que1i-300x303.png 300w\" sizes=\"auto, (max-width: 683px) 100vw, 683px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From the graph, it is can be seen that the given lines cross each other at point (3, 5).<br \/>\nSo, the cost of a pencil is 3\/- and cost of a pen is 5\/-.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. On comparing the ratios a<sub>1<\/sub>\/a<sub>2<\/sub> , b<sub>1<\/sub>\/b<sub>2<\/sub> , c<sub>1<\/sub>\/c<sub>2<\/sub>, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:<br \/>\n<\/strong><strong>(i) 5x \u2013 4y + 8 = 0<br \/>\n<\/strong><strong>7x + 6y \u2013 9 = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) 9x + 3y + 12 = 0<br \/>\n<\/strong><strong>18x + 6y + 24 = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) 6x \u2013 3y + 10 = 0<br \/>\n<\/strong><strong>2x \u2013 y + 9 = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) 5x \u2013 4y + 8 = 0<br \/>\n7x + 6y \u2013 9 = 0<br \/>\n<\/strong>Comparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 5, b<sub>1<\/sub> = -4, c<sub>1<\/sub> = 8<br \/>\na<sub>2<\/sub> = 7, b<sub>2<\/sub> = 6, c<sub>2<\/sub> = -9<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 5\/7<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -4\/6 = -2\/3<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = 8\/-9<br \/>\nSince, (a<sub>1<\/sub>\/a<sub>2<\/sub>) \u2260 (b<sub>1<\/sub>\/b<sub>2<\/sub>)<br \/>\nSo, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) 9x + 3y + 12 = 0<br \/>\n<\/strong><strong>18x + 6y + 24 = 0<br \/>\n<\/strong>Comparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 9, b<sub>1<\/sub> = 3, c<sub>1<\/sub> = 12<br \/>\na<sub>2<\/sub> = 18, b<sub>2<\/sub> = 6, c<sub>2<\/sub> = 24<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 9\/18 = 1\/2<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 3\/6 = 1\/2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = 12\/24 = 1\/2<br \/>\nSince (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) = (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nSo, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) 6x \u2013 3y + 10 = 0<br \/>\n<\/strong><strong>2x \u2013 y + 9 = 0<br \/>\n<\/strong>Comparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 6, b<sub>1<\/sub> = -3, c<sub>1<\/sub> = 10<br \/>\na<sub>2<\/sub> = 2, b<sub>2<\/sub> = -1, c<sub>2<\/sub> = 9<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 6\/2 = 3\/1<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -3\/-1 = 3\/1<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = 10\/9<br \/>\nSince (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) \u2260 (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nSo, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point, and there is no possible solution for the given pair of equations.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. On comparing the ratio, (a<sub>1<\/sub>\/a<sub>2<\/sub>) , (b<sub>1<\/sub>\/b<sub>2<\/sub>) , (c<sub>1<\/sub>\/c<sub>2<\/sub>), find out whether the following pair of linear equations are consistent, or inconsistent.<br \/>\n<\/strong><strong>(i) 3x + 2y = 5 ; 2x \u2013 3y = 7<br \/>\n<\/strong><strong>(ii) 2x \u2013 3y = 8 ; 4x \u2013 6y = 9<br \/>\n<\/strong><strong>(iii) (3\/2)x+(5\/3)y = 7; 9x \u2013 10y = 14<br \/>\n<\/strong><strong>(iv) 5x \u2013 3y = 11 ; \u2013 10x + 6y = \u201322<br \/>\n<\/strong><strong>(v) (4\/3)x+2y = 8 ; 2x + 3y = 12<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) 3x + 2y = 5 ; 2x \u2013 3y = 7<br \/>\n<\/strong>3x + 2y &#8211; 5 = 0;<br \/>\n2x \u2013 3y -7 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 3, b<sub>1<\/sub> = 2, c<sub>1<\/sub> = -5<br \/>\na<sub>2<\/sub> = 2, b<sub>2<\/sub> = -3, c<sub>2<\/sub> = -7<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 3\/2<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 2\/-3<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -5\/-7 = 5\/7<br \/>\nSince, (a<sub>1<\/sub>\/a<sub>2<\/sub>) \u2260 (b<sub>1<\/sub>\/b<sub>2<\/sub>)<br \/>\nSo, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) 2x \u2013 3y = 8 ; 4x \u2013 6y = 9<br \/>\n<\/strong>2x &#8211; 3y &#8211; 8 = 0;<br \/>\n4x &#8211; 6y &#8211; 9 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 2, b<sub>1<\/sub> = -3, c<sub>1<\/sub> = -8<br \/>\na<sub>2<\/sub> = 4, b<sub>2<\/sub> = -6, c<sub>2<\/sub> = -9<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/4 = 1\/2<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -3\/-6 = 1\/2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -8\/-9 = 8\/9<br \/>\nSince , (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) \u2260 (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nSo, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) (3\/2)x+(5\/3)y = 7; 9x \u2013 10y = 14<br \/>\n<\/strong>(3\/2)x + (5\/3)y &#8211; 7 = 0;<br \/>\n9x \u2013 10y &#8211; 14 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 3\/2, b<sub>1<\/sub> = 5\/3, c<sub>1<\/sub> = -7<br \/>\na<sub>2<\/sub> = 9, b<sub>2<\/sub> = -10, c<sub>2<\/sub> = -14<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 3\/(2\u00d79) = 1\/6<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 5\/(3\u00d7 -10)= -1\/6<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -7\/-14 = 1\/2<br \/>\nSince, (a<sub>1<\/sub>\/a<sub>2<\/sub>) \u2260 (b<sub>1<\/sub>\/b<sub>2<\/sub>)<br \/>\nSo, the equations are intersecting\u00a0 each other at one point and they have only one possible solution. Hence, the equations are consistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) 5x \u2013 3y = 11 ; \u2013 10x + 6y = \u201322<br \/>\n<\/strong>5x \u2013 3y &#8211; 11 = 0;<br \/>\n\u2013 10x + 6y + 22 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 5, b<sub>1<\/sub> = -3, c<sub>1<\/sub> = -11<br \/>\na<sub>2<\/sub> = -10, b<sub>2<\/sub> = 6, c<sub>2<\/sub> = 22<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 5\/(-10) = -5\/10 = -1\/2<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -3\/6 = -1\/2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -11\/22 = -1\/2<br \/>\nSince (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) = (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nThese linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v) (4\/3)x+2y = 8 ; 2x + 3y = 12<br \/>\n<\/strong>(4\/3)x + 2y &#8211; 8 = 0;<br \/>\n2x + 3y &#8211; 12 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 4\/3 , b<sub>1 <\/sub>= 2 , c<sub>1<\/sub> = -8<br \/>\na<sub>2<\/sub> = 2, b<sub>2<\/sub> = 3 , c<sub>2<\/sub> = -12<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 4\/(3\u00d72)= 4\/6 = 2\/3<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 2\/3<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -8\/-12 = 2\/3<br \/>\nSince (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) = (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nThese linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Which of the following pairs of linear equations are consistent\/inconsistent? If consistent, obtain the solution graphically:<br \/>\n<\/strong><strong>(i) x + y = 5, 2x + 2y = 10<br \/>\n<\/strong><strong>(ii) x \u2013 y = 8, 3x \u2013 3y = 16<br \/>\n<\/strong><strong>(iii) 2x + y \u2013 6 = 0, 4x \u2013 2y \u2013 4 = 0<br \/>\n<\/strong><strong>(iv) 2x \u2013 2y \u2013 2 = 0, 4x \u2013 4y \u2013 5 = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) x + y = 5, 2x + 2y = 10<br \/>\n<\/strong>x + y &#8211; 5 = 0,<br \/>\n2x + 2y &#8211; 10 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 1 , b<sub>1 <\/sub>= 1 , c<sub>1<\/sub> = &#8211; 5<br \/>\na<sub>2<\/sub> = 2, b<sub>2<\/sub> = 2 , c<sub>2<\/sub> = -10<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 1\/2<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 1\/2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -5\/-10 = 1\/2<br \/>\nSince (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) = (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\n\u2234 The equations are coincident and they have infinite number of possible solutions.<br \/>\nSo, the equations are consistent.<br \/>\nFor, x + y = 5<br \/>\nx = 5 \u2013 y<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse; height: 43px;\" border=\"1\">\n<tbody>\n<tr style=\"height: 19px;\">\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">For 2x + 2y = 10<br \/>\nx = (10 &#8211; 2y)\/2 = 5 &#8211; y<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse; height: 43px;\" border=\"1\">\n<tbody>\n<tr style=\"height: 19px;\">\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%; height: 19px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"width: 25%; height: 24px;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">So, the equations are represented in graphs as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6179\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"735\" height=\"761\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-i.png 735w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-i-290x300.png 290w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-i-300x311.png 300w\" sizes=\"auto, (max-width: 735px) 100vw, 735px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From the figure, we can see, that the lines are overlapping each other. Therefore, the equations have infinite possible solutions.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) x \u2013 y = 8, 3x \u2013 3y = 16<br \/>\n<\/strong>x \u2013 y &#8211; 8 = 0,<br \/>\n3x \u2013 3y &#8211; 16 = 0<br \/>\nComparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 1 , b<sub>1 <\/sub>= &#8211; 1 , c<sub>1<\/sub> = &#8211; 8<br \/>\na<sub>2<\/sub> = 3, b<sub>2<\/sub> = -3 , c<sub>2<\/sub> = &#8211; 16<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 1\/3<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -1\/-3 = 1\/3<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = 8\/16 = 1\/2<br \/>\nSince, (a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) \u2260 (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nThe equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) 2x + y \u2013 6 = 0, 4x \u2013 2y \u2013 4 = 0<br \/>\n<\/strong>Comparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 2 , b<sub>1 <\/sub>= 1 , c<sub>1<\/sub> = &#8211; 6<br \/>\na<sub>2<\/sub> = 4, b<sub>2<\/sub> = -2 , c<sub>2<\/sub> = &#8211; 4<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/4 = \u00bd<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 1\/-2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -6\/-4 = 3\/2<br \/>\nSince, (a<sub>1<\/sub>\/a<sub>2<\/sub>) \u2260 (b<sub>1<\/sub>\/b<sub>2<\/sub>)<br \/>\nThe given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.<br \/>\nNow, for 2x + y \u2013 6 = 0<br \/>\ny = 6 \u2013 2x<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">6<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">And for 4x \u2013 2y \u2013 4 = 0<br \/>\ny = (4x &#8211; 4)\/2 = 2x &#8211; 2<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">So, the equations are represented in graphs as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6180\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-iii.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"635\" height=\"679\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-iii.png 635w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-iii-281x300.png 281w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que4-iii-300x321.png 300w\" sizes=\"auto, (max-width: 635px) 100vw, 635px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From the graph, it can be seen that these lines are intersecting each other at only one point (2,2).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) 2x \u2013 2y \u2013 2 = 0, 4x \u2013 4y \u2013 5 = 0<br \/>\n<\/strong>Comparing these equations with a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0<br \/>\nAnd a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0<br \/>\nWe get,<br \/>\na<sub>1<\/sub> = 2 , b<sub>1 <\/sub>= -2 , c<sub>1<\/sub> = &#8211; 2<br \/>\na<sub>2<\/sub> = 4, b<sub>2<\/sub> = -4 , c<sub>2<\/sub> = &#8211; 5<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/4 = \u00bd<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = -2\/-4 = 1\/2<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = 2\/5<br \/>\nSince, a<sub>1<\/sub>\/a<sub>2<\/sub> = b<sub>1<\/sub>\/b<sub>2<\/sub> \u2260 c<sub>1<\/sub>\/c<sub>2<br \/>\n<\/sub>Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<br \/>\n<\/strong>Let the width of the garden be <i>x<\/i> and length be <i>y<\/i>.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, according to the question, we can express the given condition as;<br \/>\ny \u2013 x = 4<br \/>\nand<br \/>\ny + x = 36<br \/>\nNow, taking y \u2013 x = 4<br \/>\ny = x + 4<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">8<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">12<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">12<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">16<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">For y + x = 36,<br \/>\ny = 36 \u2013 x<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">36<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">16<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">36<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">20<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The graphical representation of both the equation is as follows;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6181\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"695\" height=\"733\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que5.png 695w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que5-284x300.png 284w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que5-300x316.png 300w\" sizes=\"auto, (max-width: 695px) 100vw, 695px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From the graph you can see, the lines intersect each other at a point (16, 20). Hence, the width of the garden is 16 and length is 20.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. Given the linear equation 2x + 3y \u2013 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:<br \/>\n<\/strong><strong>(i) Intersecting lines<br \/>\n<\/strong><strong>(ii) Parallel lines<br \/>\n<\/strong><strong>(iii) Coincident lines<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) Given the linear equation 2x + 3y \u2013 8 = 0.<br \/>\n<\/strong>To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy the below condition;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) \u2260 (b<sub>1<\/sub>\/b<sub>2<\/sub>)<br \/>\nThus, another equation could be 2x \u2013 7y + 9 = 0, such that;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/2 = 1 and (b<sub>1<\/sub>\/b<sub>2<\/sub>) = 3\/-7<br \/>\nClearly, you can see another equation satisfies the condition.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) Given the linear equation 2x + 3y \u2013 8 = 0.<br \/>\n<\/strong>To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy the below condition;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) \u2260 (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nThus, another equation could be 6x + 9y + 9 = 0, such that;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/6 = 1\/3<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 3\/9= 1\/3<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -8\/9<br \/>\nClearly, you can see another equation satisfies the condition.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) Given the linear equation 2x + 3y \u2013 8 = 0.<br \/>\n<\/strong>To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy the below condition;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = (b<sub>1<\/sub>\/b<sub>2<\/sub>) = (c<sub>1<\/sub>\/c<sub>2<\/sub>)<br \/>\nThus, another equation could be 4x + 6y \u2013 16 = 0, such that;<br \/>\n(a<sub>1<\/sub>\/a<sub>2<\/sub>) = 2\/4 = 1\/2 ,<br \/>\n(b<sub>1<\/sub>\/b<sub>2<\/sub>) = 3\/6 = 1\/2,<br \/>\n(c<sub>1<\/sub>\/c<sub>2<\/sub>) = -8\/-16 = 1\/2<br \/>\nClearly, you can see another equation satisfies the condition.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Draw the graphs of the equations x \u2013 y + 1 = 0 and 3x + 2y \u2013 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Given, the equations for graphs are x \u2013 y + 1 = 0 and 3x + 2y \u2013 12 = 0.<br \/>\nFor, x \u2013 y + 1 = 0<br \/>\nx = -1 + y<br \/>\n<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">1<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">For, 3x + 2y \u2013 12 = 0<br \/>\nx = (12 &#8211; 2y)\/3<\/span><\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">y<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">6<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">x<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">2<\/span><\/td>\n<td style=\"width: 25%;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, the graphical representation of these equations is as follows;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6182\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"713\" height=\"739\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que7.png 713w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que7-289x300.png 289w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.2-Que7-300x311.png 300w\" sizes=\"auto, (max-width: 713px) 100vw, 713px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (\u22121, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (\u22121, 0), and (4, 0).<\/span><\/p>\n<p><span style=\"font-family: Georgia, Palatino;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 3 (Pair of Linear Equations in Two Variables)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 3 Pair of Linear Equations in Two Variables Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 3 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1058,1060,1044,1049,1048],"class_list":["post-6019","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-3-pair-of-linear-equations-in-two-variables-solutions","tag-ncert-class-10-mathematics-exercise-3-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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3 (Pair of Linear Equations in Two Variables)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 3 Pair of Linear Equations in Two Variables Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 3.2\u00a0","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-2\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2","og_description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 3 (Pair of Linear Equations in Two Variables)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 3 Pair of Linear Equations in Two Variables Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 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