{"id":6023,"date":"2023-07-23T05:38:32","date_gmt":"2023-07-23T05:38:32","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6023"},"modified":"2023-07-23T05:44:42","modified_gmt":"2023-07-23T05:44:42","slug":"ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-6","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-6\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6"},"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 3.6 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-3-7\/\" 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; 3.6<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Solve the following pairs of equations by reducing them to a pair of linear equations:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6210\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.6-Que1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"548\" height=\"458\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.6-Que1.png 548w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex3.6-Que1-300x251.png 300w\" sizes=\"auto, (max-width: 548px) 100vw, 548px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; \u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{2x}&amp;space;+&amp;space;\\frac{1}{3y}}\" alt=\"\\mathbf{\\frac{1}{2x} + \\frac{1}{3y}}\" align=\"absmiddle\" \/> = 2 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{3x}+\\frac{1}{2y}&amp;space;=&amp;space;\\frac{13}{6}}\" alt=\"\\mathbf{\\frac{1}{3x}+\\frac{1}{2y} = \\frac{13}{6}}\" align=\"absmiddle\" \/><br \/>\nLet us assume 1\/x = m and 1\/y = n, then the equation will change as follows.<br \/>\nm\/2 + n\/3 = 2<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3m&amp;space;+&amp;space;2n}{6}&amp;space;=&amp;space;2\" alt=\"\\frac{3m + 2n}{6} = 2\" align=\"absmiddle\" \/><br \/>\n3m + 2n &#8211; 12 = 0\u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026. (i)<br \/>\nm\/3 + n\/2 = 13\/6<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2m&amp;space;+&amp;space;3n}{6}\" alt=\"\\frac{2m + 3n}{6}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{13}{6}\" alt=\"\\frac{13}{6}\" align=\"absmiddle\" \/><br \/>\n2m + 3n &#8211; 13 = 0\u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026. (ii)<br \/>\nNow, using the cross-multiplication method, we get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{m}{(-26&amp;space;-(-36))}&amp;space;=&amp;space;\\frac{n}{(-24-(-39))}&amp;space;=&amp;space;\\frac{1}{9-4}\" alt=\"\\frac{m}{(-26 -(-36))} = \\frac{n}{(-24-(-39))} = \\frac{1}{9-4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">m\/10 = n\/15 = 1\/5<br \/>\nm\/10 = 1\/5 and n\/15 = 1\/5<br \/>\nSo, m = 2 and n = 3<br \/>\n1\/x = 2 and 1\/y = 3<br \/>\nx = 1\/2 and y = 1\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{2}{\\sqrt{x}}&amp;space;+&amp;space;\\frac{3}{\\sqrt{y}}}&amp;space;=&amp;space;2\" alt=\"\\mathbf{\\frac{2}{\\sqrt{x}} + \\frac{3}{\\sqrt{y}}} = 2\" align=\"absmiddle\" \/><br \/>\n<\/strong><\/span><\/p>\n<p><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{4}{\\sqrt{x}}&amp;space;+&amp;space;\\frac{9}{\\sqrt{y}}}&amp;space;=&amp;space;-1\" alt=\"\\mathbf{\\frac{4}{\\sqrt{x}} + \\frac{9}{\\sqrt{y}}} = -1\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Substituting 1\/\u221ax = m and 1\/\u221ay = n in the given equations, we get<br \/>\n2m + 3n = 2\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\n4m \u2013 9n = -1\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026 (ii)<br \/>\nMultiplying equation (i) by 3, we get<br \/>\n6m + 9n = 6\u00a0 &#8230;.\u2026\u2026.\u2026.. (iii)<br \/>\nAdding equations (ii) and (iii), we get<br \/>\n10m = 5<br \/>\nm = 1\/2\u00a0 \u00a0\u2026\u2026\u2026.\u2026 (iv)<br \/>\nNow, by putting the value of \u2018m\u2019 in equation (i), we get<br \/>\n2\u00d7 \u00bd + 3n = 2<br \/>\n3n = 1<br \/>\nn =\u00a01\/3<br \/>\nm =1\/\u221ax<br \/>\n\u00bd = 1\/\u221ax<br \/>\nx = 4<br \/>\nn = 1\/\u221ay<br \/>\n1\/3 = 1\/\u221ay<br \/>\ny = 9<br \/>\nHence, x = 4 and y = 9<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{4}{x}&amp;space;+&amp;space;3y&amp;space;=&amp;space;14}\" alt=\"\\mathbf{\\frac{4}{x} + 3y = 14}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3}{x}&amp;space;-&amp;space;4y&amp;space;=&amp;space;23}\" alt=\"\\mathbf{\\frac{3}{x} - 4y = 23}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Substituting 1\/x = m\u00a0 in the given equation, we get,<br \/>\n4m + 3y = 14<br \/>\n4m + 3y \u2013 14 = 0\u00a0 \u2026\u2026\u2026\u2026\u2026&#8230; (i)<br \/>\n3m \u2013 4y = 23<br \/>\n3m \u2013 4y \u2013 23 = 0\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<br \/>\nBy cross-multiplication, we get,<br \/>\nm\/(-69 &#8211; 56) = y\/(-42 -(-92)) = 1\/(-16-9)<br \/>\n-m\/125 = y\/50 = -1\/ 25<br \/>\n-m\/125 = -1\/25 and y\/50 = -1\/25<br \/>\nm = 5 and b = -2<br \/>\nm =\u00a01\/x\u00a0= 5<br \/>\nSo ,<br \/>\nx = 1\/5<br \/>\ny = -2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{5}{(x-1)}+\\frac{1}{(y-2)}=&amp;space;2}\" alt=\"\\mathbf{\\frac{5}{(x-1)}+\\frac{1}{(y-2)}= 2}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{6}{(x-1)}-\\frac{3}{(y-2)}=&amp;space;1}\" alt=\"\\mathbf{\\frac{6}{(x-1)}-\\frac{3}{(y-2)}= 1}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Substituting 1\/(x &#8211; 1) = m and 1\/(y &#8211; 2) = n \u00a0in the given equations, we get<br \/>\n5m + n = 2\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026 (i)<br \/>\n6m \u2013 3n = 1\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (ii)<br \/>\nMultiplying equation (i) by 3, we get<br \/>\n15m + 3n = 6\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (iii)<br \/>\nAdding (ii) and (iii), we get<br \/>\n21m = 7<br \/>\nm =\u00a01\/3<br \/>\nPutting this value in equation (i), we get<br \/>\n5 \u00d7 1\/3 + n = 2<br \/>\nn = 2 &#8211; 5\/3 = 1\/3<br \/>\nm = 1\/ (x &#8211; 1)<br \/>\n1\/3 = 1\/(x &#8211; 1)<br \/>\nx = 4<br \/>\nn = 1\/(y &#8211; 2)<br \/>\n1\/3 = 1\/(y &#8211; 2)<br \/>\ny = 5<br \/>\nHence, x = 4 and y = 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{7x&amp;space;-&amp;space;2y}{xy}&amp;space;=&amp;space;5}\" alt=\"\\mathbf{\\frac{7x - 2y}{xy} = 5}\" width=\"101\" height=\"41\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{8x&amp;space;+&amp;space;7y}{xy}&amp;space;=&amp;space;15}\" alt=\"\\mathbf{\\frac{8x + 7y}{xy} = 15}\" width=\"110\" height=\"41\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7x&amp;space;-&amp;space;2y}{xy}&amp;space;=&amp;space;5\" alt=\"\\frac{7x - 2y}{xy} = 5\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">7\/y \u2013 2\/x = 5\u00a0 \u2026\u2026\u2026\u2026.. (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{8x&amp;space;+&amp;space;7y}{xy}&amp;space;=&amp;space;15\" alt=\"\\frac{8x + 7y}{xy} = 15\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">8\/y + 7\/x = 15\u00a0 \u2026\u2026\u2026\u2026\u2026 (ii)<br \/>\nSubstituting 1\/x =m in the given equation, we get<br \/>\n\u2013 2m + 7n = 5<br \/>\n-2 + 7n \u2013 5 = 0\u00a0 \u2026\u2026.. (iii)<br \/>\n7m + 8n = 15<br \/>\n7m + 8n \u2013 15 = 0 \u2026\u2026.. (iv)<br \/>\nBy cross-multiplication method, we get<br \/>\nm\/(-105 &#8211; (-40)) = n\/(-35 &#8211; 30) = 1\/(-16 &#8211; 49)<br \/>\nm\/(-65) = n\/(-65) = 1\/(-65)<br \/>\nm\/-65 = 1\/-65<br \/>\nm = 1<br \/>\nn\/(-65) = 1\/(-65)<br \/>\nn = 1<br \/>\nm = 1 and n = 1<br \/>\nm =\u00a01\/x\u00a0= 1<br \/>\nn =\u00a01\/x\u00a0= 1<br \/>\nTherefore, x = 1 and y = 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) 6x + 3y = 6xy<br \/>\n<\/strong><strong>2x + 4y = 5xy<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>6x + 3y = 6xy<br \/>\n6\/y + 3\/x = 6<br \/>\nLet 1\/x = m and 1\/y = n<br \/>\n6n +3m = 6<br \/>\n3m + 6n &#8211; 6 = 0\u00a0 \u2026\u2026\u2026\u2026. (i)<br \/>\n<strong>2x + 4y = 5xy<br \/>\n<\/strong>2\/y + 4\/x = 5<br \/>\n2n + 4m = 5<br \/>\n4m + 2n &#8211; 5 = 0\u00a0 \u2026\u2026\u2026\u2026.. (ii)<br \/>\n3m + 6n \u2013 6 = 0<br \/>\n4m + 2n \u2013 5 = 0<br \/>\nBy cross-multiplication method, we get<br \/>\nm\/(-30 \u2013(-12)) = n\/(-24-(-15)) = 1\/(6-24)<br \/>\nm\/-18 = n\/-9 = 1\/-18<br \/>\nm\/-18 = 1\/-18<br \/>\nm = 1<br \/>\nn\/-9 = 1\/-18<br \/>\nn = 1\/2<br \/>\nm = 1 and n = 1\/2<br \/>\nm = 1\/x = 1 and n = 1\/y = 1\/2<br \/>\nx = 1\u00a0and\u00a0y = 2<br \/>\nHence, x = 1 and y = 2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{10}{(x&amp;space;+&amp;space;y)}&amp;space;+&amp;space;\\frac{2}{(x-y)}&amp;space;=&amp;space;4}\" alt=\"\\mathbf{\\frac{10}{(x + y)} + \\frac{2}{(x-y)} = 4}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{15}{(x&amp;space;+&amp;space;y)}&amp;space;+&amp;space;\\frac{5}{(x-y)}&amp;space;=&amp;space;-2}\" alt=\"\\mathbf{\\frac{15}{(x + y)} + \\frac{5}{(x-y)} = -2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Substituting 1\/(x + y) = m and 1\/(x &#8211; y) = n in the given equations, we get<br \/>\n10m + 2n = 4<br \/>\n10m + 2n \u2013 4 = 0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (i)<br \/>\n15m \u2013 5n = -2<br \/>\n15m \u2013 5n + 2 = 0\u00a0 \u00a0 \u2026\u2026\u2026\u2026&#8230;.. (ii)<br \/>\nUsing the cross-multiplication method, we get<br \/>\nm\/(4 &#8211; 20) = n\/(-60 &#8211; (20)) = 1\/(-50 -30)<br \/>\nm\/-16 = n\/-80 = 1\/-80<br \/>\nm\/-16 = 1\/-80 and n\/-80 = 1\/-80<br \/>\nm = 1\/5 and n = 1<br \/>\nm = 1\/(x + y) = 1\/5<br \/>\nx + y = 5\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<br \/>\nn = 1\/(x &#8211; y) = 1<br \/>\nx &#8211; y = 1 \u2026\u2026\u2026\u2026\u2026\u2026 (iv)<br \/>\nAdding equations (iii) and (iv), we get<br \/>\n2x = 6<br \/>\nx = 3 \u2026\u2026. (v)<br \/>\nPutting the value of x = 3 in equation (3), we get<br \/>\ny = 2<br \/>\nHence, x = 3 and y = 2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(viii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{(3x+y)}&amp;space;+&amp;space;\\frac{1}{3x&amp;space;-&amp;space;y}&amp;space;=&amp;space;\\frac{3}{4}}\" alt=\"\\mathbf{\\frac{1}{(3x+y)} + \\frac{1}{3x - y} = \\frac{3}{4}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{2(3x+y)}&amp;space;+&amp;space;\\frac{1}{2(3x&amp;space;-&amp;space;y)}&amp;space;=&amp;space;-\\frac{1}{8}}\" alt=\"\\mathbf{\\frac{1}{2(3x+y)} + \\frac{1}{2(3x - y)} = -\\frac{1}{8}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Substituting 1\/(3x + y) = m and 1\/(3x &#8211; y) = n in the given equations, we get<br \/>\nm + n = 3\/4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026.\u2026\u2026 (i)<br \/>\nm\/2 \u2013 n\/2 = -1\/8<br \/>\nm \u2013 n = -1\/4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026..\u2026 (ii)<br \/>\nAdding (i) and (ii), we get<br \/>\n2m = 3\/4 \u2013 1\/4<br \/>\n2m = 1\/2<br \/>\nPutting in (ii), we get<br \/>\n1\/4 \u2013 n = -1\/4<br \/>\nn = 1\/4 + 1\/4 = 1\/2<br \/>\nm = 1\/(3x + y) = 1\/4<br \/>\n3x + y = 4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)<br \/>\nn = 1\/(3x &#8211; y) = 1\/2<br \/>\n3x \u2013 y = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<br \/>\nAdding equations (iii) and (iv), we get<br \/>\n6x = 6<br \/>\nx = 1\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (5)<br \/>\nPutting in (iii), we get<br \/>\n3(1) + y = 4<br \/>\ny = 1<br \/>\nHence, x = 1 and y = 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Formulate the following problems as a pair of equations and find their solutions.<br \/>\n<\/strong><strong>(i) Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong>Let us consider,<br \/>\nSpeed of Ritu in still water = x km\/hr<br \/>\nSpeed of Stream = y km\/hr<br \/>\nNow, the speed of Ritu during<br \/>\nDownstream = x + y km\/h<br \/>\nUpstream = x \u2013 y km\/h<br \/>\nAs per the question given,<br \/>\n2(x + y) = 20<br \/>\nx + y = 10 \u2026\u2026\u2026\u2026\u2026\u2026. (i)<br \/>\n2 (x &#8211; y) = 4<br \/>\nx \u2013 y = 2 \u2026\u2026\u2026\u2026\u2026 (2)<br \/>\nAdding both the eq. (i) and (ii), we get<br \/>\n2x = 12<br \/>\nx = 6<br \/>\nPutting the value of x in eq. (i), we get<br \/>\ny = 4<br \/>\nTherefore,<br \/>\nSpeed of Ritu rowing in still water = 6 km\/hr<br \/>\nSpeed of Stream = 4 km\/hr<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work and also that taken by 1 man alone.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong>Let us consider,<br \/>\nNumber of days taken by women to finish the work = x<br \/>\nNumber of days taken by men to finish the work = y<br \/>\nWork done by women in one day = 1\/x<br \/>\nWork done by women in one day = 1\/y<br \/>\nAs per the question given,<br \/>\n4(2\/x + 5\/y) = 1<br \/>\n(2\/x + 5\/y) = 1\/4<br \/>\n(3\/x + 6\/y) = 1<br \/>\n(3\/x + 6\/y) = 1\/3<br \/>\nNow, put 1\/x=m and 1\/y=n, we get,<br \/>\n2m + 5n = 1\/4<br \/>\n8m + 20n = 1\u00a0 \u00a0\u2026\u2026\u2026 (i)<br \/>\n3m + 6n = 1\/3<br \/>\n9m + 18n = 1 \u2026\u2026. (ii)<br \/>\nNow, by cross multiplication method, we get here,<br \/>\nm\/(20 &#8211; 18) = n\/(9 &#8211; 8) = 1\/ (180 &#8211; 144)<br \/>\nm\/2 = n\/1 = 1\/36<br \/>\nm\/2 = 1\/36<br \/>\nm = 1\/18<br \/>\nm = 1\/x = 1\/18<br \/>\nx = 18<br \/>\nn = 1\/y = 1\/36<br \/>\ny = 36<br \/>\nTherefore,<br \/>\nNumber of days taken by women to finish the work = 18<br \/>\nNumber of days taken by men to finish the work = 36<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong>Let us consider,<br \/>\nSpeed of the train = x km\/h<br \/>\nSpeed of the bus = y km\/h<br \/>\nAccording to the given question,<br \/>\n60\/x + 240\/y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026 (i)<br \/>\n100\/x + 200\/y = 25\/6\u00a0 \u00a0 \u2026\u2026\u2026. (ii)<br \/>\nPut 1\/x = m and 1\/y = n in the above two equations.<br \/>\n60m + 240n = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026&#8230;. (iii)<br \/>\n100m + 200n = 25\/6<br \/>\n600m + 1200n = 25\u00a0 \u00a0 \u00a0\u2026\u2026\u2026&#8230; (iv)<br \/>\nMultiply eq. (iii) by 10 to get<br \/>\n600m + 2400n = 40\u00a0 ..\u2026\u2026\u2026\u2026 (v)<br \/>\nNow, subtract eq. (iv) from 5 to get<br \/>\n1200n = 15<br \/>\nn = 15\/1200 = 1\/80<br \/>\nSubstitute the value of n in eq. (iii) to get<br \/>\n60m + 3 = 4<br \/>\nm = 1\/60<br \/>\nm = 1\/x = 1\/60<br \/>\nx = 60<br \/>\ny = 1\/n<br \/>\ny = 80<br \/>\nTherefore,<br \/>\nSpeed of the train = 60 km\/h<br \/>\nSpeed of the bus = 80 km\/h<\/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 3.6 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":[1045,1058,1064,1044,1049,1048],"class_list":["post-6023","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-3-6-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 v27.3) - 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