{"id":11113,"date":"2022-06-20T17:00:27","date_gmt":"2022-06-20T11:30:27","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11113"},"modified":"2022-06-20T17:00:30","modified_gmt":"2022-06-20T11:30:30","slug":"which-of-the-following-pairs-of-linear-equations-has-unique-solutions-no-solution-or-infinitely-solution-in-case-there-is-a-unique-solution-find-it-by-using-cross-multiplication-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/which-of-the-following-pairs-of-linear-equations-has-unique-solutions-no-solution-or-infinitely-solution-in-case-there-is-a-unique-solution-find-it-by-using-cross-multiplication-method\/","title":{"rendered":"Which of the following pairs of linear equations has unique solutions, no solution, or infinitely solution ? In case there is a unique solution, find it by using cross multiplication method."},"content":{"rendered":"
Question<\/strong> : Which of the following pairs of linear equations has unique solutions, no solution, or infinitely solution ? In case there is a unique solution, find it by using cross multiplication method.<\/p>\n (i)<\/strong>\u00a0 x – 3y – 3 = 0\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0 3x – 9y – 2 = 0<\/p>\n (ii)<\/strong>\u00a0 2x + y = 5\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 \u00a03x + 2y = 8<\/p>\n (iii)<\/strong>\u00a0 3x – 5y = 20\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 6x – 10y = 40<\/p>\n (iv)<\/strong>\u00a0 \u00a0x – 3y – 7 = 0\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0 3x – 3y – 15 = 0<\/p>\n (i)<\/strong>\u00a0 The given equations are<\/p>\n x – 3y – 3 = 0<\/p>\n and\u00a0 3x – 9y – 2 = 0<\/p>\n These equations are of the form\u00a0 \u00a0\\(a_1x + b_1y +\u00a0 c_1\\) = 0<\/p>\n and\u00a0 \\(a_2x + b_2y + c_2\\) = 0<\/p>\n where \\(a_1\\) = 1, \\(b_1\\) = -3, \\(c_1\\) = -3\u00a0 \u00a0and\u00a0 \u00a0\\(a_2\\) = 3, \\(b_2\\) = -9,\u00a0 \\(c_2\\) = -2<\/p>\n We have :\u00a0 \\(a_1\\over a_2\\) = \\(1\\over 3\\),\u00a0 \\(b_1\\over b_2\\) = \\(-3\\over -9\\) = \\(1\\over 3\\)<\/p>\n and\u00a0 \\(c_1\\over c_2\\)\u00a0 =\u00a0 \\(-3\\over -2\\) = \\(3\\over 2\\)<\/p>\n Clearly we see that, \\(a_1\\over a_2\\) = \\(b_1\\over b_2\\) \\(\\ne\\)\u00a0 \\(c_1\\over c_2\\)<\/p>\n So, the given linear system of equations has no solution (i.e. system of equations is inconsistent)<\/strong>.<\/p>\n (ii)<\/strong>\u00a0The given equations are<\/p>\n 2x + y – 5 = 0<\/p>\n and\u00a0 3x + 2y – 8 = 0<\/p>\n These equations are of the form\u00a0 \u00a0\\(a_1x + b_1y +\u00a0 c_1\\) = 0<\/p>\n and\u00a0 \\(a_2x + b_2y + c_2\\) = 0<\/p>\n where \\(a_1\\) = 2, \\(b_1\\) = 1, \\(c_1\\) = -5\u00a0 \u00a0and\u00a0 \u00a0\\(a_2\\) = 3, \\(b_2\\) = 2,\u00a0 \\(c_2\\) = -8<\/p>\n We have :\u00a0 \\(a_1\\over a_2\\) = \\(2\\over 3\\),\u00a0 \\(b_1\\over b_2\\) = \\(1\\over 2\\)<\/p>\n Clearly we see that, \\(a_1\\over a_2\\) \\(\\ne\\) \\(b_1\\over b_2\\)<\/p>\n So, the given linear system of equations has unique solution.<\/strong><\/p>\n To find the solution, we will be using cross-multiplication method here. By cross-multiplication method<\/strong>,<\/p>\n \\(x\\over {1\\times (-8) – 2\\times (-5)}\\) = \\(y\\over {-5 \\times 3 – (-8) \\times 2}\\) = \\(1\\over {2\\times 2 – 3\\times 1}\\)<\/p>\n \\(\\implies\\)\u00a0 \u00a0\\(x\\over -8 + 10\\) = \\(y\\over -15 + 16\\) = \\(1\\over 4 – 3\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over 2\\) = \\(y\\over 1\\) = \\(1\\over 1\\)\u00a0 \u00a0 \u00a0or\u00a0 \u00a0 x = 2,\u00a0 y = 1<\/p>\n Hence, the given linear system of equation has unique solution given by x = 2, y = 1<\/strong>.<\/p>\n (iii)<\/strong>\u00a0 The given equations are<\/p>\n 3x – 5y – 20 = 0<\/p>\n and\u00a0 6x – 10y – 40 = 0<\/p>\n These equations are of the form\u00a0 \u00a0\\(a_1x + b_1y +\u00a0 c_1\\) = 0<\/p>\n and\u00a0 \\(a_2x + b_2y + c_2\\) = 0<\/p>\n where \\(a_1\\) = 3, \\(b_1\\) = -5, \\(c_1\\) = -20\u00a0 \u00a0and\u00a0 \u00a0\\(a_2\\) = 6, \\(b_2\\) = -10,\u00a0 \\(c_2\\) = -40<\/p>\n We have :\u00a0 \\(a_1\\over a_2\\) = \\(3\\over 6\\) = \\(1\\over 2\\),\u00a0 \\(b_1\\over b_2\\) = \\(-5\\over -10\\) = \\(1\\over 2\\)<\/p>\n and\u00a0 \\(c_1\\over c_2\\)\u00a0 =\u00a0 \\(-20\\over -40\\) = \\(1\\over 2\\)<\/p>\n Clearly we see that, \\(a_1\\over a_2\\) = \\(b_1\\over b_2\\) = \\(c_1\\over c_2\\)<\/p>\n So, the given linear system of equations has infinitely many solutions.<\/strong><\/p>\n (iv)<\/strong>\u00a0The given equations are<\/p>\n x – 3y – 7 = 0<\/p>\n and\u00a0 3x – 3y – 15 = 0<\/p>\n These equations are of the form\u00a0 \u00a0\\(a_1x + b_1y +\u00a0 c_1\\) = 0<\/p>\n and\u00a0 \\(a_2x + b_2y + c_2\\) = 0<\/p>\n where \\(a_1\\) = 1, \\(b_1\\) = -3, \\(c_1\\) = -7\u00a0 \u00a0and\u00a0 \u00a0\\(a_2\\) = 3, \\(b_2\\) = -3,\u00a0 \\(c_2\\) = -15<\/p>\n We have :\u00a0 \\(a_1\\over a_2\\) = \\(1\\over 3\\),\u00a0 \\(b_1\\over b_2\\) = 1<\/p>\n Clearly we see that, \\(a_1\\over a_2\\) \\(\\ne\\) \\(b_1\\over b_2\\)<\/p>\n So, the given linear system of equations has unique solution.<\/strong><\/p>\n To find the solution, we will be using cross-multiplication method here. By cross-multiplication method<\/strong>,<\/p>\n \u00a0\\(x\\over 45 – 21\\) = \\(y\\over -21 + 15\\) = \\(1\\over -3 + 9\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over 24\\) = \\(y\\over -6\\) = \\(1\\over 6\\)\u00a0 \u00a0 \u00a0or\u00a0 \u00a0 x = 4,\u00a0 y = -1<\/p>\n Hence, the given linear system of equation has unique solution given by x = 4, y = -1<\/strong>.<\/p>\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Question : Which of the following pairs of linear equations has unique solutions, no solution, or infinitely solution ? In case there is a unique solution, find it by using cross multiplication method. (i)\u00a0 x – 3y – 3 = 0\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0 3x – 9y – 2 = 0 (ii)\u00a0 2x …<\/p>\nSolution :<\/h2>\n