{"id":6024,"date":"2023-07-23T05:38:46","date_gmt":"2023-07-23T05:38:46","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6024"},"modified":"2023-07-23T05:44:28","modified_gmt":"2023-07-23T05:44:28","slug":"ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-7","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-7\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 3 (Pair of Linear Equations in Two Variables)\u00a0<\/span><\/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 3.7 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-3-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 3.2<\/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<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 3.7<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1.\u00a0<\/strong><b>The ages of two friends Ani and Biju differ by 3 years. Ani\u2019s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let the age of Ani and Biju be x and y years respectively.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the age of Ani\u2019s father, Dharam is 2x years,<\/span><br \/>\n<span style=\"color: #000000;\">And the age of Biju\u2019s sister Cathy is <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> years.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case (I) :<\/strong>\u00a0When Ani is older than Biju<\/span><br \/>\n<span style=\"color: #000000;\">The ages of Ani and Biju differ by 3 years,<\/span><br \/>\n<span style=\"color: #000000;\">x &#8211; y = 3\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">The ages of Cathy and Dharam differs by 30 years,<\/span><br \/>\n<span style=\"color: #000000;\">2x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = 30<\/span><br \/>\n<span style=\"color: #000000;\">4x &#8211; y = 60 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting (i) from (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">4x &#8211; y &#8211; (x &#8211; y) = 60 &#8211; 3<\/span><br \/>\n<span style=\"color: #000000;\">3x = 57<\/span><br \/>\n<span style=\"color: #000000;\">x = 19<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 19 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">19 &#8211; y = 3<\/span><br \/>\n<span style=\"color: #000000;\">y = 16<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Ani is 19 years old and Biju is 16 years old.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Case (II) :<\/strong>\u00a0When Biju is older than Ani.<\/span><br \/>\n<span style=\"color: #000000;\">The ages of Ani and Biju differ by 3 years,<\/span><br \/>\n<span style=\"color: #000000;\">y &#8211; x = 3 <\/span><br \/>\n<span style=\"color: #000000;\">&#8211; x + y = 3\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">The ages of Cathy and Dharam differs by 30 years,<\/span><br \/>\n<span style=\"color: #000000;\">2x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = 30<\/span><br \/>\n<span style=\"color: #000000;\">4x &#8211; y = 60 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Adding (i) and (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; x + y +\u00a04x &#8211; y = 3 + 60<\/span><br \/>\n<span style=\"color: #000000;\">3x = 63<\/span><br \/>\n<span style=\"color: #000000;\">x = 21<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 21 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 21 + y = 3<\/span><br \/>\n<span style=\"color: #000000;\">y = 24<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Ani is 21 years old and Biju is 24 years old.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, Ani is 19 years old and Biju is 16 years old or Ani is 21 years old and Biju is 24 years old.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2.\u00a0<b>One says, \u201cGive me a hundred, friend! I shall then become twice as rich as you\u201d. The other replies, \u201cIf you give me ten, I shall be six times as rich as you\u201d. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]<br \/>\n[Hint : x + 100 = 2(y \u2013 100), y + 10 = 6(x \u2013 10)].<\/b><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let the first friend have \u20b9 x<\/span><br \/>\n<span style=\"color: #000000;\">And the second friend has \u20b9 y<\/span><br \/>\n<span style=\"color: #000000;\">Using the information given in the question,<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Condition 1:<\/strong>\u00a0When second friend gives \u20b9 100 to first friend;<\/span><br \/>\n<span style=\"color: #000000;\">x + 100 = 2 (y &#8211; 100)<\/span><br \/>\n<span style=\"color: #000000;\">x + 100 = 2y &#8211; 200<\/span><br \/>\n<span style=\"color: #000000;\">x &#8211; 2y = &#8211; 300\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Condition 2:<\/strong> When first friend gives \u20b9 10 to second friend; <\/span><br \/>\n<span style=\"color: #000000;\">y + 10 = 6 (x &#8211; 10)<\/span><br \/>\n<span style=\"color: #000000;\">y + 10 = 6x &#8211; 60<\/span><br \/>\n<span style=\"color: #000000;\">6x &#8211; y = 70\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Multiplying equation (ii) by 2, we obtain<\/span><br \/>\n<span style=\"color: #000000;\">12x &#8211; 2y = 140\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (iii)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting equation (i) from equation (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">12x &#8211; 2y &#8211; (x &#8211; 2y) = 140 &#8211; (- 300)<\/span><br \/>\n<span style=\"color: #000000;\">11x = 440<\/span><br \/>\n<span style=\"color: #000000;\">x = 440\/11<\/span><br \/>\n<span style=\"color: #000000;\">x = 40<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 40 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">40 &#8211; 2y = &#8211; 300<\/span><br \/>\n<span style=\"color: #000000;\">2y = 40 + 300<\/span><br \/>\n<span style=\"color: #000000;\">y = 340\/2<\/span><br \/>\n<span style=\"color: #000000;\">y = 170<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the first friend has \u20b9 40, and the second friend has \u20b9 170 with them.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3.\u00a0<\/strong><b>A train covered a certain distance at a uniform speed. If the train would have been 10 km\/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km\/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let us assume the uniform speed of the train to be x km\/h and the time taken to travel the\u00a0given distance be t hours.<\/span><br \/>\n<span style=\"color: #000000;\">Then distance can be calculated as follows:<\/span><br \/>\n<span style=\"color: #000000;\">Distance = speed \u00d7 time = <em>xt<br \/>\n<\/em>Thus, the distance is <em>xt<br \/>\n<\/em>According to the question,<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Condition 1:<\/strong>\u00a0When the train would have been 10 km\/h faster, it would have taken 2 hours less than the scheduled time.<\/span><br \/>\n<span style=\"color: #000000;\">(x + 10)(t &#8211; 2) = xt<\/span><br \/>\n<span style=\"color: #000000;\">xt &#8211; 2x + 10t &#8211; 20 = xt<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 2x + 10t = 20\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Condition 2:<\/strong>\u00a0When the train would have been slower by 10 km\/h, it would have taken 3 hours more than the scheduled time.<\/span><br \/>\n<span style=\"color: #000000;\">(x &#8211; 10)(t + 3) = xt<\/span><br \/>\n<span style=\"color: #000000;\">xt + 3x &#8211; 10t &#8211; 30 = xt<\/span><br \/>\n<span style=\"color: #000000;\">3x -10t = 30\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (i) and (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 2x + 10t + 3x -10t = 20 + 30 <\/span><br \/>\n<span style=\"color: #000000;\">x = 50<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 50 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 2 \u00d7 50 + 10t = 20<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 100 + 10t = 20<\/span><br \/>\n<span style=\"color: #000000;\">10t = 120<\/span><br \/>\n<span style=\"color: #000000;\">t = 120\/10<\/span><br \/>\n<span style=\"color: #000000;\">t = 12<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, distance = xt = 50 \u00d7 12 = 600<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the distance covered by the train is 600 km.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4.\u00a0<\/strong><b>The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let us assume the number of rows to be equal to x and the number of students in each row to be y.<\/span><br \/>\n<span style=\"color: #000000;\">Then the total number of students in the class can be calculated as follows:<\/span><br \/>\n<span style=\"color: #000000;\">Total number of students = Number of rows \u00d7 Number of students in each row = <em>xy<\/em><\/span><br \/>\n<span style=\"color: #000000;\">Using the information given in the question,<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Condition 1 :<\/strong>\u00a0If 3 students are extra in a row, there would be 1 row less<\/span><br \/>\n<span style=\"color: #000000;\">(x &#8211; 1)(y + 3) = xy<\/span><br \/>\n<span style=\"color: #000000;\">xy + 3x &#8211; y &#8211; 3 = xy<\/span><br \/>\n<span style=\"color: #000000;\">3x &#8211; y = 3 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Condition 2:<\/strong>\u00a0If 3 students are less in a row, there would be 2 rows more<\/span><br \/>\n<span style=\"color: #000000;\">(x + 2)(y &#8211; 3) = xy<\/span><br \/>\n<span style=\"color: #000000;\">xy &#8211; 3x + 2y &#8211; 6 = xy<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 3x + 2y = 6\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (i) and (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">3x &#8211; y + (- 3x + 2y) = 3 + 6<\/span><br \/>\n<span style=\"color: #000000;\">y = 9<\/span><br \/>\n<span style=\"color: #000000;\">Substituting y = 9 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">3x &#8211; 9 = 3<\/span><br \/>\n<span style=\"color: #000000;\">3x = 12<\/span><br \/>\n<span style=\"color: #000000;\">x = 4<\/span><br \/>\n<span style=\"color: #000000;\">Hence, number of students in the class, xy = 4 \u00d7 9 = 36<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5.\u00a0<\/strong><b>In a \u2206ABC, \u2220 C = 3 \u2220 B = 2 (\u2220 A + \u2220 B). Find the three angles.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let the measurement of \u2220A = x<\/span><br \/>\n<span style=\"color: #000000;\">And the measurement of \u2220B = y<\/span><br \/>\n<span style=\"color: #000000;\">Using the information given in the question,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220C = 3\u2220B = 2 (\u2220A + \u2220B)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 3\u2220B = 2 (\u2220A + \u2220B)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 3y = 2 ( x + y )<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 3y = 2x + 2 y<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 2x &#8211; y = 0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">We know that the sum of the measures of all angles of a triangle is 180\u00b0. Therefore,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220B + \u2220C = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220A + \u2220B + 3\u2220B = 180\u00b0\u00a0[\u2235 \u2220C = 3\u2220B]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220A + 4\u2220B = 180<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x + 4y = 180\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Multiplying equation (i) by 4, we obtain<\/span><br \/>\n<span style=\"color: #000000;\">8x &#8211; 4y = 0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (iii)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (ii) and (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">x + 4y +\u00a08x &#8211; 4y = 180 + 0<\/span><br \/>\n<span style=\"color: #000000;\">9x = 180<\/span><br \/>\n<span style=\"color: #000000;\">x = 20<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 20 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">2 \u00d7 20 &#8211; y = 0<\/span><br \/>\n<span style=\"color: #000000;\">y = 40<\/span><br \/>\n<span style=\"color: #000000;\">Therefore,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = x = 20\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220B = y = 40\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220C = 3\u2220B = 3 \u00d7 40\u00b0 = 120\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6.\u00a0<\/strong><b>Draw the graphs of the equations 5x \u2013 y = 5 and 3x \u2013 y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions: <\/strong>Given,<br \/>\n5x \u2013 y = 5<br \/>\n\u21d2\u00a0y = 5x \u2013 5<br \/>\nThe solution table will be as follows.<\/span><\/p>\n<table border=\"1\">\n<tbody>\n<tr>\n<td width=\"81\"><span style=\"color: #000000;\">x<\/span><\/td>\n<td width=\"84\"><span style=\"color: #000000;\">0<\/span><\/td>\n<td width=\"74\"><span style=\"color: #000000;\">2<\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"81\"><span style=\"color: #000000;\">y<\/span><\/td>\n<td width=\"84\"><span style=\"color: #000000;\">-5<\/span><\/td>\n<td width=\"74\"><span style=\"color: #000000;\">5<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #000000;\">3x &#8211; y = 3<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 y = 3x &#8211; 3<\/span><br \/>\n<span style=\"color: #000000;\">The solution table will be as follows.<\/span><\/p>\n<table border=\"1\">\n<tbody>\n<tr>\n<td width=\"82\"><span style=\"color: #000000;\">x<\/span><\/td>\n<td width=\"81\"><span style=\"color: #000000;\">0<\/span><\/td>\n<td width=\"79\"><span style=\"color: #000000;\">2<\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"82\"><span style=\"color: #000000;\">y<\/span><\/td>\n<td width=\"81\"><span style=\"color: #000000;\">-3<\/span><\/td>\n<td width=\"79\"><span style=\"color: #000000;\">3<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #000000;\">The graphical representation of these lines will be as follows.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6233 size-large\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6-1024x599.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"640\" height=\"374\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6-1024x599.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6-300x176.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6-768x449.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6-850x497.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Ans6.png 1425w\" sizes=\"auto, (max-width: 640px) 100vw, 640px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">From the above graph, we can see that the coordinates of the vertices of the triangle formed by the lines and the y-axis are (1, 0), (0, -5) and (0, -3).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7.\u00a0<\/strong><b>Solve the following pair of linear equations:<br \/>\n<\/b><b>(i) <\/b>px + qy = p \u2013 q<br \/>\nqx \u2013 py = p + q<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>(ii) <\/b>ax + by = c<br \/>\nbx + ay = 1 + c<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>(iii) <\/b>x\/a \u2013 y\/b = 0<br \/>\nax + by = a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>(iv) <\/b>(a \u2013 b)x + (a + b) y = a<sup>2<\/sup>\u00a0\u2013 2ab \u2013 b<sup>2<br \/>\n<\/sup>(a + b)(x + y) = a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>(v) <\/b>152x \u2013 378y = \u2013 74<br \/>\n\u2013378x + 152y = \u2013 604<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> px + qy = p &#8211; q\u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">qx &#8211; py = p + q\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Multiplying equation (i) by p and equation (ii) by q, we obtain<\/span><br \/>\n<span style=\"color: #000000;\">p\u00b2 x + pqy = p\u00b2 &#8211; pq &#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">q\u00b2 x &#8211; pqy = pq + q\u00b2 &#8212;&#8212;&#8212;&#8211; (iv)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (iii) and (iv), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">p\u00b2x + q\u00b2x = p\u00b2 + q\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(p\u00b2 + q\u00b2) x = p\u00b2 + q\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">x = (p\u00b2 + q\u00b2)\/p\u00b2+ q\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">x = 1<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x =1 in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">p \u00d7 1 + qy = p &#8211; q<\/span><br \/>\n<span style=\"color: #000000;\">qy = &#8211; q<\/span><br \/>\n<span style=\"color: #000000;\">y = &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, x = 1 and y = &#8211; 1<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)<\/strong> ax + by = c\u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">bx + ay = 1+ c\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Multiplying equation (i) by a and equation (ii) by b , we obtain<\/span><br \/>\n<span style=\"color: #000000;\">a\u00b2x + aby = ac\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">b\u00b2x + aby = b + bc\u00a0 &#8212;&#8212;&#8212;&#8211; (iv)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting equation (iv) from equation (iii),<\/span><br \/>\n<span style=\"color: #000000;\">(a\u00b2 &#8211; b\u00b2) x = ac &#8211; bc &#8211; b<\/span><br \/>\n<span style=\"color: #000000;\">x = [c(a &#8211; b) &#8211; b]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = [c(a &#8211; b) &#8211; b]\/(a\u00b2 &#8211; b\u00b2) in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">ax + by = c<\/span><br \/>\n<span style=\"color: #000000;\">a [c(a &#8211; b) &#8211; b]\/(a\u00b2 &#8211; b\u00b2) + by\u00a0= c<\/span><br \/>\n<span style=\"color: #000000;\">[ac(a &#8211; b) &#8211; ab]\/(a\u00b2 &#8211; b\u00b2) + by = c<\/span><br \/>\n<span style=\"color: #000000;\">by = c &#8211; [ac(a &#8211; b) &#8211; ab]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">by = [a\u00b2c &#8211; b\u00b2c &#8211; a\u00b2c + abc + ab]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">by = [abc &#8211; b\u00b2c + ab]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">by = [bc(a &#8211; b) + ab]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">by = b [c(a &#8211; b) + a]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">y = [c(a &#8211; b) + a]\/(a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, x = [c(a &#8211; b) &#8211; b]\/(a\u00b2 &#8211; b\u00b2) and y = [c(a &#8211; b) + a]\/(a\u00b2 &#8211; b\u00b2)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii)<\/strong> x\/a &#8211; y\/b = 0 &#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">ax + by = a\u00b2 + b\u00b2 &#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">By solving equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">x\/a &#8211; y\/b = 0<\/span><br \/>\n<span style=\"color: #000000;\">x = ay\/b\u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = ay\/b in equation (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">a (ay\/b) + by = a\u00b2 + b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(a\u00b2 y + b\u00b2 y)\/b = a\u00b2 + b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(a\u00b2 + b\u00b2) y = b (a\u00b2 + b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">y = b<\/span><br \/>\n<span style=\"color: #000000;\">Substituting y = b in equation (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">x = (a \u00d7 b)\/b<\/span><br \/>\n<span style=\"color: #000000;\">x = a<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, x = a and y = b<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv)<\/strong> (a &#8211; b)x + (a + b) y = a\u00b2 &#8211; 2ab &#8211; b\u00b2 &#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">(a + b)(x + y) = a\u00b2 + b\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">By solving equation (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">(a + b)(x + y) = a\u00b2 + b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(a + b)x + (a + b) y = a\u00b2 + b\u00b2 &#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting equation (iii) from (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">(a &#8211; b)x &#8211; (a + b)x = (a\u00b2 &#8211; 2ab &#8211; b\u00b2) &#8211; (a\u00b2 + b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">[(a &#8211; b) &#8211; (a + b)] x = a\u00b2 &#8211; 2ab &#8211; b\u00b2 &#8211; a\u00b2 &#8211; b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">[a &#8211; b &#8211; a &#8211; b] x = -2ab &#8211; 2b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">-2bx = -2b (a + b)<\/span><br \/>\n<span style=\"color: #000000;\">x = (a + b)<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = (a + b) in equation (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">(a &#8211; b)(a + b) + (a + b) y = a\u00b2 &#8211; 2ab &#8211; b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(a\u00b2 &#8211; b\u00b2) + (a + b) y = a\u00b2 &#8211; 2ab &#8211; b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">(a + b) y = a\u00b2 &#8211; 2ab &#8211; b\u00b2 &#8211; (a\u00b2 &#8211; b\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">(a + b) y = a\u00b2 &#8211; 2ab &#8211; b\u00b2 &#8211; a\u00b2 + b\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">y = -2ab\/(a + b)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(v)<\/strong> 152x &#8211; 378y = -74 &#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; 378x + 152y = &#8211; 604 &#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (i) and (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">-226x &#8211; 226y = &#8211; 678<\/span><br \/>\n<span style=\"color: #000000;\">-226( x + y ) = &#8211; 678<\/span><br \/>\n<span style=\"color: #000000;\">x + y = 3 &#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting equation (ii) from (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">530x &#8211; 530y = 530<\/span><br \/>\n<span style=\"color: #000000;\">530 ( x &#8211; y ) = 530<\/span><br \/>\n<span style=\"color: #000000;\">x &#8211; y = 1 &#8212;&#8212;&#8212;&#8211; (iv)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equations (iii) and (iv), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">2x = 4<\/span><br \/>\n<span style=\"color: #000000;\">x = 2<\/span><br \/>\n<span style=\"color: #000000;\">Substituting x = 2 in equation (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">2 + y = 3<\/span><br \/>\n<span style=\"color: #000000;\">y = 1<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, x = 2 and y = 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6234 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Que8-270x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"270\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Que8-270x300.png 270w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Que8-300x333.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex3.7-Que8.png 390w\" sizes=\"auto, (max-width: 270px) 100vw, 270px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that ABCD is a cyclic quadrilateral.<br \/>\nAs we know, the opposite angles of a cyclic quadrilateral are supplementary.<br \/>\nSo,<br \/>\n\u2220A + \u2220C = 180<br \/>\n4y + 20 + (-4x) = 180<br \/>\n-4x + 4y = 160<br \/>\n\u21d2 -x + y = 40\u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nAnd<br \/>\n\u2220B + \u2220D = 180<br \/>\n3y \u2013 5 + (-7x + 5) = 180<br \/>\n\u21d2 -7x + 3y = 180 &#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nEquation (ii) \u2013 3 \u00d7 (i),<br \/>\n-7x + 3y \u2013 (-3x + 3y) = 180 \u2013 120<br \/>\n-4x = 60<br \/>\nx = -15<br \/>\nSubstituting x = -15 in equation (i), we get;<br \/>\n-(-15) + y = 40<br \/>\ny = 40 \u2013 15 = 25<br \/>\nTherefore, x = -15 and y = 25.<\/span><\/p>\n<p><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><\/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 3.7 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":[1066,1067,1068,1065,1048],"class_list":["post-6024","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-maths-solutions","tag-ncert-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-solutions","tag-ncert-class-10-maths-ex-3-7-solutions","tag-ncert-class-10-maths-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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