Solve the following pair of linear equations :<\/p>\n
(i)<\/strong>\u00a0 \u00a0px + qy = p – q\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 qx – py = p + q<\/p>\n (ii)<\/strong>\u00a0 ax + by = c\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 bx + ay = 1 + c<\/p>\n (iii)<\/strong>\u00a0 \\(x\\over a\\) – \\(y\\over b\\) = 0\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 ax + by = \\(a^2 + b^2\\)<\/p>\n (iv)<\/strong>\u00a0 Solve for x and y :<\/p>\n (a – b)x + (a + b)y = \\(a^2 – 2ab\u00a0 – b^2\\)<\/p>\n (a + b)(x + y) = \\(a^2 + b^2\\)<\/p>\n (v)\u00a0<\/strong> 152x – 378y = -74\u00a0 \u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 \u00a0 -378x + 152y = -604<\/p>\n (i)<\/strong>\u00a0 The given linear equations are<\/p>\n px + qy = p – q\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 px + qy – (p – q) = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(1)<\/p>\n qx – py = p + q\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0qx – py – (p + q) = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(2)<\/p>\n Solving it by cross multiplication method, we get<\/p>\n \\(x\\over -q(p + q) – p(p – q)\\) = \\(y\\over -q(p – q) + p(p + q)\\) = \\(1\\over -p^2 – q^2\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over -pq – q^2 – p^2 + pq\\) = \\(y\\over – pq + q^2 + p^2 + pq\\) = \\(1\\over -(p^2 + q^2)\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over -(p^2 + q^2)\\) = \\(y\\over p^2 + q^2\\) = \\(1\\over -(p^2 + q^2)\\)<\/p>\n \\(\\implies\\)\u00a0 x = 1 and y = 1.<\/strong><\/p>\n (ii)<\/strong>\u00a0 The given linear equations are<\/p>\n ax + by – c = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(1)<\/p>\n bx + ay – (1 + c) = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(2)<\/p>\n Solving it by cross multiplication method, we get<\/p>\n \\(x\\over -b(1 + c) + ac\\) = \\(y\\over -bc + a(1 + c)\\) = \\(1\\over a^2 – b^2\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over -b – bc + ac\\) = \\(y\\over -bc + a + ac\\) = \\(1\\over a^2 – b^2)\\)<\/p>\n \\(\\implies\\)\u00a0\u00a0\\(x\\over c(a – b) – b\\) = \\(y\\over c(a – b) + a\\) = \\(1\\over (a – b)(a + b)\\)<\/p>\n \\(\\implies\\)\u00a0 x = \\(c\\over a + b\\) – \\(b\\over (a – b)(a + b)\\)<\/strong><\/p>\n y = \\(c\\over a + b\\) + \\(b\\over (a – b)(a + b)\\)<\/strong><\/p>\n (iii)<\/strong>\u00a0 The given linear equations are<\/p>\n \\(x\\over a\\) – \\(y\\over b\\) = 0\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 bx – ay = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(1)<\/p>\n ax + by -\\((a^2 + b^2)\\) = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(2)<\/p>\n Solving it by cross multiplication method, we get<\/p>\n \\(x\\over a(a^2 + b^2) – 0\\) = \\(y\\over 0 + b(a^2 + b^2)\\) = \\(1\\over b^2 + a^2\\)<\/p>\n \\(\\implies\\)\u00a0 \u00a0\\(x\\over a(a^2 + b^2)\\) = \\(y\\over b(a^2 + b^2)\\) = \\(1\\over a^2 + b^2\\)<\/p>\n \\(\\implies\\)\u00a0 x = a,\u00a0 y = b<\/strong>.<\/p>\n (iv)<\/strong>\u00a0 The given linear equations are<\/p>\n (a – b)x + (a + b)y = \\(a^2 – 2ab\u00a0 – b^2\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(1)<\/p>\n (a + b)(x + y) = \\(a^2 + b^2\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(2)<\/p>\n Solving it by cross multiplication method, we get<\/p>\n \\(x\\over -(a + b)(a^2 + b^2) + (a + b)(a^2 – 2ab – b^2)\\) = \\(y\\over -(a + b)(a^2 – 2ab – b^2) + (a – b)(a^2 + b^2)\\) = \\(1\\over (a – b)(a + b) – {(a + b)}^2\\)<\/p>\n \\(\\implies\\)\u00a0 \u00a0\\(x\\over (a + b)(-a^2 – b^2 + a^2 – 2ab – b^2)\\) = \\(y\\over -a^3 + 2a^2b + ab^2 – a^2b + 2ab^2 – b^3 + a^3 + ab^2 – a^2b – b^3\\) = \\(1\\over a^2 – b^2 – a^2 – b^2 – 2ab\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over (a + b)(-2ab – 2b^2)\\) = \\(y\\over 4ab^2\\) = \\(1\\over -2b^2 – 2ab\\)<\/p>\n \\(\\implies\\)\u00a0 \\(x\\over (a + b)(-2b)(a + b)\\) = \\(y\\over 4ab^2\\) = \\(1\\over -2b(a + b)\\)<\/p>\nSolution :<\/h2>\n