{"id":10987,"date":"2022-06-11T14:46:59","date_gmt":"2022-06-11T09:16:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10987"},"modified":"2022-06-11T14:47:02","modified_gmt":"2022-06-11T09:17:02","slug":"solve-the-following-pair-of-linear-equations-by-the-substitution-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-the-following-pair-of-linear-equations-by-the-substitution-method\/","title":{"rendered":"Solve the following pair of linear equations by the substitution method :"},"content":{"rendered":"

Question<\/strong> : Solve the following pair of linear equations by the substitution method :<\/p>\n

(i)<\/strong>\u00a0 x + y = 14\u00a0 \u00a0 and\u00a0 \u00a0 \u00a0x – y = 4<\/p>\n

(ii)<\/strong> s – t = 3\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 \\(s\\over 3\\) + \\(t\\over 2\\) = 6<\/p>\n

(iii)<\/strong>\u00a0 3x – y = 3\u00a0 \u00a0 and\u00a0 \u00a0 \u00a09x – 3y = 9<\/p>\n

(iv)<\/strong>\u00a0 0.2x + 0.3y = 1.3\u00a0 \u00a0 and\u00a0 \u00a00.4x + 0.5y = 2.3<\/p>\n

(v)<\/strong>\u00a0 \\(\\sqrt{2}x + \\sqrt{3}y\\) = 0\u00a0 \u00a0 and\u00a0 \u00a0 \\(\\sqrt{3}x – \\sqrt{8}y\\) = 0<\/p>\n

(vi)<\/strong>\u00a0 \\(3x\\over 2\\) – \\(5y\\over 3\\) = 2\u00a0 \u00a0and\u00a0 \\(x\\over 3\\) + \\(y\\over 2\\) = \\(13\\over 6\\)<\/p>\n

Solution<\/strong> :<\/h2>\n

(i)<\/strong>\u00a0 The given system of equations is<\/p>\n

x + y = 14\u00a0 \u00a0 \u00a0 …….(1)<\/p>\n

and\u00a0 \u00a0 \u00a0x – y = 4\u00a0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

From equation (1),\u00a0 y = 14 – x<\/p>\n

Substituting y = 14 – x in equation (2), we get<\/p>\n

x – (14 – x) = 14\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x – 14 + x = 4<\/p>\n

\\(\\implies\\)\u00a0 2x = 4 + 14\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 2x = 18<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x = 9<\/p>\n

Putting\u00a0 \u00a0x = 9 in equation (1), we get<\/p>\n

9 + y = 14\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0y = 5<\/p>\n

Hence, the solution of the given system of linear equations is x = 9 and y = 5<\/strong>.<\/p>\n

(ii)<\/strong>\u00a0 The given system of equations is<\/p>\n

s – t = 3\u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 \u00a0 \u00a0 \\(s\\over 3\\) + \\(t\\over 2\\) = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…..(2)<\/p>\n

From equation (1),\u00a0 s = 3 + t<\/p>\n

Substituting s = 3 + t in equation (2), we get<\/p>\n

\\(3 + t\\over 3\\) + \\(t\\over 2\\) = 6 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 2(3 + t) + 3t = 36<\/p>\n

\\(\\implies\\)\u00a0 6 + 2t + 3t = 36\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 5t = 30\u00a0 \u00a0 \\(\\implies\\)\u00a0 t = 6<\/p>\n

Putting\u00a0 \u00a0t = 6 in equation (1), we get<\/p>\n

s – 6 = 3\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0s = 9<\/p>\n

Hence, the solution of the given system of linear equations is s = 3 and t = 6<\/strong>.<\/p>\n

(iii)<\/strong>\u00a0 The given system of equations is<\/p>\n

3x – y = 3\u00a0 \u00a0 \u00a0 …….(1)<\/p>\n

and\u00a0 \u00a0 \u00a09x – 3y = 9\u00a0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

From equation (1),\u00a0 y = 3x – 3<\/p>\n

Substituting y = 3x – 3 in equation (2), we get<\/p>\n

9x – 3(3x – 3) = 9\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 9x – 9x + 9 = 9<\/p>\n

or\u00a0 \u00a09 = 9<\/p>\n

Since, this statement always remains true for all value of x.<\/p>\n

Therefore, Equation (1) and (2) have infinitely many solutions.<\/strong><\/p>\n

(iv)<\/strong>\u00a0 The given system of equations is<\/p>\n

0.2x + 0.3y = 1.3\u00a0 \u00a0 \\(\\implies\\)\u00a0 2x + 3y = 13 \u00a0 \u00a0 …….(1)<\/p>\n

and\u00a0 \u00a00.4x + 0.5y = 2.3\u00a0 \\(\\implies\\)\u00a0 4x + 5y = 23\u00a0 \u00a0 \u00a0……(2)<\/p>\n

From equation (2),\u00a0 5y = 23 – 4x\u00a0 \\(\\implies\\)\u00a0 y = \\(23 – 4x\\over 5\\)<\/p>\n

Substituting y = \\(23 – 4x\\over 5\\) in equation (1), we get<\/p>\n

10x + 69 – 12x = 65\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 -2x = -4<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x = 2<\/p>\n

Putting\u00a0 \u00a0x = 2 in equation (1), we get<\/p>\n

3y = 13 – 4\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0y = 3<\/p>\n

Hence, the solution of the given system of linear equations is x = 2 and y = 3<\/strong>.<\/p>\n

(v)<\/strong>\u00a0 The given system of equations is<\/p>\n

\\(\\sqrt{2}x + \\sqrt{3}y\\) = 0\u00a0 \u00a0 \u00a0\u00a0 \u00a0…….(1)<\/p>\n

and\u00a0 \u00a0 \\(\\sqrt{3}x – \\sqrt{8}y\\) = 0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

From equation (2),\u00a0 y = \\(\\sqrt{3}x\\over \\sqrt{8}\\)<\/p>\n

Substituting y = \\(\\sqrt{3}x\\over \\sqrt{8}\\) in equation (1), we get<\/p>\n

\\(\\sqrt{2}x + \\sqrt{3y}({\\sqrt{3}x\\over \\sqrt{8}})\\) = 0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0\\(\\sqrt{2}x\\) + \\(3x\\over \\sqrt{8}\\) = 0<\/p>\n

\\(\\implies\\)\u00a0 4x + 3x = 0\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 7x = 0<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x = 0<\/p>\n

Putting\u00a0 \u00a0x = 0 in equation (1), we get<\/p>\n

0 + \\(\\sqrt{3}y\\) = 14\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0y = 0<\/p>\n

Hence, the solution of the given system of linear equations is x = 0 and y = 0<\/strong>.<\/p>\n

(vi)<\/strong>\u00a0 The given system of equations is<\/p>\n

\\(3x\\over 2\\) – \\(5y\\over 3\\) = 2\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 9x – 10y = -12\u00a0 ……(1)<\/p>\n

and\u00a0 \\(x\\over 3\\) + \\(y\\over 2\\) = \\(13\\over 6\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 2x + 3y = 13\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

From equation (1),\u00a0 10y = 9x + 12\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0y =\\(9x + 12\\over 10\\)<\/p>\n

Substituting y =\\(9x + 12\\over 10\\) in equation (2), we get<\/p>\n

20x + 27x + 36 = 130\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 47x = 130 – 36<\/p>\n

\\(\\implies\\)\u00a0 47x = 94\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 x = 2<\/p>\n

Putting\u00a0 \u00a0x = 2 in equation (1), we get<\/p>\n

-10y = -12 – 18\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0-10y = -30<\/p>\n

Hence, the solution of the given system of linear equations is x = 2 and y = 3<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Solve the following pair of linear equations by the substitution method : (i)\u00a0 x + y = 14\u00a0 \u00a0 and\u00a0 \u00a0 \u00a0x – y = 4 (ii) s – t = 3\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 \\(s\\over 3\\) + \\(t\\over 2\\) = 6 (iii)\u00a0 3x – y = 3\u00a0 \u00a0 and\u00a0 \u00a0 …<\/p>\n

Solve the following pair of linear equations by the substitution method :<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[911,43],"tags":[],"yoast_head":"\nSolve the following pair of linear equations by the substitution method : - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/solve-the-following-pair-of-linear-equations-by-the-substitution-method\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solve the following pair of linear equations by the substitution method : - 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