{"id":6020,"date":"2023-07-22T07:22:50","date_gmt":"2023-07-22T07:22:50","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6020"},"modified":"2023-07-23T05:45:28","modified_gmt":"2023-07-23T05:45:28","slug":"ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-ex-3-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3"},"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 3.3 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-3-4\/\" 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; 3.3\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Solve the following pair of linear equations by the substitution method.<br \/>\n<\/strong><strong>(i) x + y = 14<br \/>\n<\/strong><strong>x \u2013 y = 4<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) s \u2013 t = 3<br \/>\n<\/strong><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{s}{3}}\" alt=\"\\mathbf{\\frac{s}{3}}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{t}{2}}\" alt=\"\\mathbf{\\frac{t}{2}}\" align=\"absmiddle\" \/> = 6<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong> (iii) 3x \u2013 y = 3<br \/>\n<\/strong><strong>9x \u2013 3y = 9<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 0.2x + 0.3y = 1.3<br \/>\n<\/strong><strong>0.4x + 0.5y = 2.3<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) \u221a2x + \u221a3y = 0<br \/>\n<\/strong><strong>\u221a3x &#8211; \u221a8y = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong> (vi) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3x}{2}}\" alt=\"\\mathbf{\\frac{3x}{2}}\" align=\"absmiddle\" \/> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{5y}{3}}\" alt=\"\\mathbf{\\frac{5y}{3}}\" align=\"absmiddle\" \/> = -2<\/strong><\/span><\/p>\n<p><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{x}{3}}\" alt=\"\\mathbf{\\frac{x}{3}}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{y}{2}}\" alt=\"\\mathbf{\\frac{y}{2}}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{13}{6}}\" alt=\"\\mathbf{\\frac{13}{6}}\" align=\"absmiddle\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) x + y = 14\u00a0 \u00a0 \u00a0 \u00a0<\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<strong><br \/>\n<\/strong><strong>x \u2013 y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<strong><br \/>\n<\/strong>From the (i) equation, we get,<br \/>\nx = 14 \u2013 y<br \/>\nNow, substitute the value of x to the second equation to get,<br \/>\n(14 \u2013 y) \u2013 y = 4<br \/>\n14 \u2013 2y = 4<br \/>\n2y = 10<br \/>\ny = 10\/2<br \/>\ny = 5<br \/>\nBy the value of y, we can now find the exact value of x.<br \/>\n\u2235 x = 14 \u2013 y<br \/>\n\u2234 x = 14 \u2013 5<br \/>\nx = 9<br \/>\nHence, x = 9 and y = 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) s \u2013 t = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<strong><br \/>\n<\/strong><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{s}{3}}\" alt=\"\\mathbf{\\frac{s}{3}}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{t}{2}}\" alt=\"\\mathbf{\\frac{t}{2}}\" align=\"absmiddle\" \/> = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nFrom the (i) equation, we get,<br \/>\ns = 3 + t<br \/>\nNow, substitute the value of s to the second equation to get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3+t}{3}&amp;space;+&amp;space;\\frac{t}{2}\" alt=\"\\frac{3+t}{3} + \\frac{t}{2}\" align=\"absmiddle\" \/> \u00a0= 6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2(3+t)&amp;space;+&amp;space;3(t))}{6}\" alt=\"\\frac{2(3+t) + 3(t))}{6}\" align=\"absmiddle\" \/> = 6<br \/>\n(6 + 2t + 3t)\u00a0 = 6 \u00d7 6<br \/>\n(6 + 5t) = 36<br \/>\n5t = 30<br \/>\nt = 6<br \/>\nNow, substitute the value of t to the equation (1)<br \/>\ns \u2013 t = 3<br \/>\ns \u2013 6 = 3<br \/>\ns = 3 + 6<br \/>\ns = 9<br \/>\nTherefore, s = 9 and t = 6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 3x \u2013 y = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<strong><br \/>\n<\/strong><strong>9x \u2013 3y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nFrom the 1<sup>st<\/sup> equation, we get,<br \/>\nx = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(3+y)}{3}\" alt=\"\\frac{(3+y)}{3}\" align=\"absmiddle\" \/><br \/>\nNow, substitute the value of x to the second equation to get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{(3+y)}{3}&amp;space;-\" alt=\"9\\frac{(3+y)}{3} -\" align=\"absmiddle\" \/> 3y = 9<br \/>\n3(3 + y) &#8211; 3y = 9<br \/>\n9 + 3y &#8211; 3y = 9<br \/>\n9 = 9<br \/>\nTherefore, y has infinite values, and since x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(3+y)}{3}\" alt=\"\\frac{(3+y)}{3}\" align=\"absmiddle\" \/>, x also has infinite values.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 0.2x + 0.3y = 1.3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<strong><br \/>\n<\/strong><strong>0.4x + 0.5y = 2.3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nFrom the (i) equation, we get,<br \/>\nx = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(1.3-0.3y)}{0.2}\" alt=\"\\frac{(1.3-0.3y)}{0.2}\" align=\"absmiddle\" \/><br \/>\nNow, substitute the value of x to the second equation to get,<br \/>\n0.4 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(1.3-0.3y)}{0.2}\" alt=\"\\frac{(1.3-0.3y)}{0.2}\" align=\"absmiddle\" \/> + 0.5y = 2.3<br \/>\n2(1.3 \u2013 0.3y) + 0.5y = 2.3<br \/>\n2.6 \u2013 0.6y + 0.5y = 2.3<br \/>\n2.6 \u2013 0.1 y = 2.3<br \/>\n0.1 y = 0.3<br \/>\ny = 3<br \/>\nNow, substitute the value of y in equation (i), and we get,<br \/>\n0.2x + 0.3y = 1.3<br \/>\n0.2x + 0.3 \u00d7 3 = 1.3<br \/>\n0.2x\u00a0 = 1.3 &#8211; 0.9<br \/>\n0.2x = 0.4<br \/>\nx = 0.4\/0.2<br \/>\nx = 2<br \/>\nTherefore, x = 2 and y = 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) \u221a2x + \u221a3y = 0\u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<strong><br \/>\n<\/strong><strong>\u221a3x &#8211; \u221a8y = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nFrom the (i) equation, we get,<br \/>\nx = \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{3}}{\\sqrt{2}}y\" alt=\"\\frac{\\sqrt{3}}{\\sqrt{2}}y\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (iii)<br \/>\nPutting the value of x in the given (ii) equation to get,<br \/>\n\u221a3<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(-&amp;space;\\frac{\\sqrt{3}}{\\sqrt{2}}&amp;space;\\right&amp;space;)\" alt=\"\\left (- \\frac{\\sqrt{3}}{\\sqrt{2}} \\right )\" align=\"absmiddle\" \/> y \u2013 \u221a8y = 0<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(-&amp;space;\\frac{3}{\\sqrt{2}}&amp;space;\\right&amp;space;)\" alt=\"\\left (- \\frac{3}{\\sqrt{2}} \\right )\" align=\"absmiddle\" \/> y &#8211; 2\u221a2 y = 0<br \/>\ny <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;-\\frac{3}{\\sqrt{2}}&amp;space;-&amp;space;2\\sqrt{2}\\right&amp;space;)\" alt=\"\\left ( -\\frac{3}{\\sqrt{2}} - 2\\sqrt{2}\\right )\" align=\"absmiddle\" \/>\u00a0 = 0<br \/>\ny = 0<br \/>\nNow, substitute the value of y in equation (iii), and we get,<br \/>\nx = 0<br \/>\nTherefore, x = 0 and y = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3x}{2}}\" alt=\"\\mathbf{\\frac{3x}{2}}\" align=\"absmiddle\" \/> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{5y}{3}}\" alt=\"\\mathbf{\\frac{5y}{3}}\" align=\"absmiddle\" \/> = -2\u00a0 \u00a0 \u00a0 \u00a0 <\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (i)<\/span><\/p>\n<p><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{x}{3}}\" alt=\"\\mathbf{\\frac{x}{3}}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{y}{2}}\" alt=\"\\mathbf{\\frac{y}{2}}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{13}{6}}\" alt=\"\\mathbf{\\frac{13}{6}}\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/strong>&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nFrom (i) equation, we get,<br \/>\n<strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3x}{2}}\" alt=\"\\mathbf{\\frac{3x}{2}}\" align=\"absmiddle\" \/> <\/strong>= -2 + <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{5y}{3}}\" alt=\"\\mathbf{\\frac{5y}{3}}\" align=\"absmiddle\" \/><br \/>\n<\/strong>x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2(-6+5y)}{9}\" alt=\"\\frac{2(-6+5y)}{9}\" align=\"absmiddle\" \/><br \/>\nx = (-12 + 10y)\/9\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)<br \/>\nPutting the value of x in the (ii) equation, we get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\frac{(-12+10y)}{9}}{3}&amp;space;+&amp;space;\\frac{y}{2}\" alt=\"\\frac{\\frac{(-12+10y)}{9}}{3} + \\frac{y}{2}\" align=\"absmiddle\" \/> = 13\/6<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{13}{6}&amp;space;-&amp;space;\\frac{(-12&amp;space;+&amp;space;10y)}{27}\" alt=\"\\frac{13}{6} - \\frac{(-12 + 10y)}{27}\" align=\"absmiddle\" \/><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{13\\times&amp;space;9&amp;space;-2(-12+10y)}{54}\" alt=\"\\frac{13\\times 9 -2(-12+10y)}{54}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{117&amp;space;+&amp;space;24&amp;space;-&amp;space;20y}{54}\" alt=\"\\frac{117 + 24 - 20y}{54}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{2}\" alt=\"\\frac{y}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{141&amp;space;-20y}{54}\" alt=\"\\frac{141 -20y}{54}\" align=\"absmiddle\" \/><br \/>\n27y = 141 &#8211; 20y<br \/>\n<span style=\"color: #000000;\">27y + 20y = 141<\/span><br \/>\n<span style=\"color: #000000;\">47y = 141<\/span><br \/>\n<span style=\"color: #000000;\">y = 141\/47<\/span><br \/>\n<span style=\"color: #000000;\">y = 3<\/span><br \/>\n<span style=\"color: #000000;\">Now, substitute the value of y in equation (iii), and we get,<br \/>\nx = (-12 + 10y)\/9\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)<br \/>\nx = (-12 + 10\u00d73)\/9<br \/>\nx = (-12 + 30)\/9<br \/>\nx = 18\/9<br \/>\nx = 2<br \/>\nTherefore, x = 2 and y = 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Solve 2x + 3y = 11 and 2x \u2013 4y = \u2013 24 and hence find the value of \u2018m\u2019 for which y = mx + 3.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>2x + 3y = 11\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)<br \/>\n2x \u2013 4y = -24\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)<br \/>\nFrom equation (i), we get<br \/>\nx = (11 &#8211; 3y)\/2\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iii)<br \/>\nSubstituting the value of x to equation (ii), we get<br \/>\n2(11-3y)\/2 \u2013 4y = 24<br \/>\n11 \u2013 3y \u2013 4y = -24<br \/>\n-7y = -35<br \/>\ny = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<br \/>\nPutting the value of y in equation (iii), we get<br \/>\nx = (11 &#8211; 3\u00d75)\/2<br \/>\nx = (11 &#8211; 15)\/2<br \/>\nx = -4\/2<br \/>\nx = -2<br \/>\nHence, x = -2, y = 5<br \/>\nAlso,<br \/>\ny = mx + 3<br \/>\n5 = -2m +3<br \/>\n-2m = 2<br \/>\nm = -1<br \/>\nTherefore, the value of m is -1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Form the pair of linear equations for the following problems and find their solution by the substitution method.<br \/>\n<\/strong><strong>(i) The difference between two numbers is 26, and one number is three times the other. Find them.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let the two numbers be x and y, respectively, such that y &gt; x.<br \/>\nAccording to the question,<br \/>\ny = 3x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026 (i)<br \/>\ny \u2013 x = 26\u00a0 \u00a0 \u00a0&#8230;.\u2026\u2026\u2026\u2026.. (ii)<br \/>\nSubstituting the value of (i) to (ii), we get<br \/>\n3x \u2013 x = 26<br \/>\nx = 13\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026. (iii)<br \/>\nSubstituting (iii) in (i), we get<br \/>\ny = 3x<br \/>\ny = 3 \u00d7 13<br \/>\ny = 39<br \/>\nHence, the numbers are 13 and 39.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) The larger of two supplementary angles exceeds, the smaller by 18 degrees. Find them.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the larger angle be x<sup>o<\/sup> and the smaller angle be y<sup>o<\/sup>.<br \/>\nWe know that the sum of two supplementary pairs of angles is always 180<sup>o<\/sup>.<br \/>\nAccording to the question,<br \/>\nx + y = 180<sup>o\u00a0 \u00a0 \u00a0<\/sup>\u2026\u2026\u2026\u2026\u2026. (i)<br \/>\nx \u2013 y = 18<sup>o\u00a0 \u00a0 \u00a0 <\/sup>\u2026\u2026\u2026\u2026\u2026..(ii)<br \/>\nFrom (i), we get<br \/>\nx = 180<sup>o<\/sup> \u2013 y\u00a0 \u00a0 \u2026\u2026\u2026\u2026. (iii)<br \/>\nSubstituting (iii) in (ii), we get<br \/>\n180<sup>o\u00a0<\/sup>\u2013 y \u2013 y =18<sup>o<br \/>\n<\/sup>162<sup>o<\/sup>\u00a0= 2y<br \/>\ny = 81<sup>o<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026.. (iv)<br \/>\nUsing the value of y in (iii), we get<br \/>\nx = 180<sup>o<\/sup>\u00a0\u2013 81<sup>o<br \/>\n<\/sup>= 99<sup>o<br \/>\n<\/sup>Hence, the angles are 99<sup>o<\/sup>\u00a0and 81<sup>o<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the cost of a bat be x and the cost of a ball be y.<br \/>\nAccording to the question,<br \/>\n7x + 6y = 3800\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026. (i)<br \/>\n3x + 5y = 1750\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (ii)<br \/>\nFrom (i), we get<br \/>\ny = (3800 &#8211; 7x)\/6\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026.. (iv)<br \/>\nSubstituting (iii) to (ii), we get,<br \/>\n3x + 5<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{3800&amp;space;-&amp;space;7x}{6}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\frac{3800 - 7x}{6} \\right )\" align=\"absmiddle\" \/> = 1750<\/span><\/p>\n<p>3x + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9500}{3}\" alt=\"\\frac{9500}{3}\" align=\"absmiddle\" \/> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{35x}{6}\" alt=\"\\frac{35x}{6}\" align=\"absmiddle\" \/> = 1750<\/p>\n<p>3x \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{35x}{6}\" alt=\"\\frac{35x}{6}\" align=\"absmiddle\" \/>\u00a0 = 1750 \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9500}{3}\" alt=\"\\frac{9500}{3}\" align=\"absmiddle\" \/><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">(18x &#8211; 35x)\/6 = (5250 \u2013 9500)\/3<br \/>\n-17x\/6 = -4250\/3<br \/>\n-17x = -8500<br \/>\nx = 500\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iv)<br \/>\nSubstituting the value of x to (iii), we get<br \/>\ny = (3800 &#8211; 7 \u00d7 500)\/6<br \/>\ny = 300\/6<br \/>\ny = 50<br \/>\nHence, the cost of a bat is Rs 500, and the cost of a ball is Rs 50.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the fixed charge be Rs x and the per km charge be Rs y.<br \/>\nAccording to the question,<br \/>\nx + 10y = 105\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\nx + 15y = 155\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nFrom (1), we get<br \/>\nx = 105 \u2013 10y\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026. (iii)<br \/>\nSubstituting the value of x to (ii), we get<br \/>\n105 \u2013 10y + 15y = 155<br \/>\n5y = 50<br \/>\ny = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026.. (iv)<br \/>\nPutting the value of y in (3), we get<br \/>\nx = 105 \u2013 10 \u00d7 10<br \/>\nx = 5<br \/>\nHence, the fixed charge is Rs 5 and the per km charge = Rs 10<br \/>\nCharge for 25 km = x + 25y<br \/>\n= 5 + 250<br \/>\n= Rs 255<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) A fraction becomes 9\/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5\/6. Find the fraction.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the fraction be x\/y.<br \/>\nAccording to the question,<br \/>\n(x + 2) \/ (y + 2) = 9\/11<br \/>\n11x + 22 = 9y + 18<br \/>\n11x \u2013 9y = -4\u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\n(x+3) \/(y+3) = 5\/6<br \/>\n6x + 18 = 5y +15<br \/>\n6x \u2013 5y = -3\u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<br \/>\nFrom (i), we get<br \/>\nx = (-4 + 9y)\/11 \u2026\u2026\u2026\u2026\u2026.. (iii)<br \/>\nSubstituting the value of x to (ii), we get<br \/>\n6(-4 + 9y)\/11 &#8211; 5y = -3<br \/>\n-24 + 54y \u2013 55y = -33<br \/>\n-y = -9<br \/>\ny = 9\u00a0 \u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<br \/>\nSubstituting the value of y to (iii), we get<br \/>\nx = (-4 + 9 \u00d7 9 )\/11<br \/>\nx = 7<br \/>\nHence, the fraction is 7\/9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob\u2019s age was seven times that of his son. What are their present ages?<br \/>\n<\/strong><strong>Solutions &#8211;<br \/>\n<\/strong>Let the age of Jacob and his son be x and y, respectively.<br \/>\nAccording to the question,<br \/>\n(x + 5) = 3(y + 5)<br \/>\nx \u2013 3y = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)<br \/>\n(x \u2013 5) = 7(y \u2013 5)<br \/>\nx \u2013 7y = -30\u00a0 \u00a0 \u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<br \/>\nFrom (1), we get<br \/>\nx = 3y + 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)<br \/>\nSubstituting the value of x to (ii), we get<br \/>\n3y + 10 \u2013 7y = -30<br \/>\n-4y = -40<br \/>\ny = 10\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iv)<br \/>\nSubstituting the value of y to (iii), we get<br \/>\nx = 3 \u00d7 10 + 10<br \/>\nx = 40<\/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 3.3 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":[1045,1058,1061,1044,1049,1048],"class_list":["post-6020","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-3-3-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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