{"id":11197,"date":"2022-06-26T16:17:59","date_gmt":"2022-06-26T10:47:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11197"},"modified":"2022-06-26T16:36:33","modified_gmt":"2022-06-26T11:06:33","slug":"solve-the-following-pairs-of-equations-by-reducing-them-to-a-pair-of-linear-of-linear-equations","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-the-following-pairs-of-equations-by-reducing-them-to-a-pair-of-linear-of-linear-equations\/","title":{"rendered":"Solve the following pairs of equations by reducing them to a pair of linear of linear equations"},"content":{"rendered":"

Question :<\/h2>\n

Solve the following pairs of equations by reducing them to a pair of linear of linear equations :<\/p>\n

(i)<\/strong>\u00a0 \\(1\\over 2x\\) + \\(1\\over 3y\\) = 2\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \\(1\\over 3x\\) + \\(1\\over 2y\\) = 2<\/p>\n

(ii)<\/strong>\u00a0 \\(2\\over \\sqrt{x}\\) + \\(3\\over \\sqrt{y}\\) = 2\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \\(4\\over \\sqrt{x}\\) – \\(9\\over \\sqrt{y}\\) = 2<\/p>\n

(iii)<\/strong>\u00a0 \\(4\\over x\\) + 3y = 14\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0\\(3\\over x\\) – 4y = 23<\/p>\n

(iv)<\/strong>\u00a0 \\(4\\over x\\) + 3y = 14\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0\\(6\\over x – 1\\) – \\(3\\over y – 2\\) = 1<\/p>\n

(v)<\/strong>\u00a0 \\(7x – 2y\\over xy\\)\u00a0 = 5\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0\\(8x + 7y\\over xy\\)\u00a0 =\u00a0 15<\/p>\n

(vi)<\/strong>\u00a0 6x + 3y = 6xy\u00a0 \u00a0 \u00a0 and\u00a0 \u00a0 \u00a0 \u00a02x + 4y = 5xy<\/p>\n

(vii)<\/strong>\u00a0 \\(10\\over x + y\\) + \\(2\\over x – y\\) = 4\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \u00a0\\(15\\over x + y\\) – \\(5\\over x – y\\) = -2<\/p>\n

(viii)<\/strong>\u00a0 \\(1\\over 3x + y\\) + \\(1\\over 3x – y\\) = \\(3\\over 4\\)\u00a0 \u00a0and\u00a0 \u00a0\\(1\\over 2(3x + y)\\) – \\(1\\over 2(3x – y)\\) = \\(-1\\over 8\\)<\/p>\n

Solution :<\/h2>\n

(i)<\/strong>\u00a0 Put \\(1\\over x\\) = u\u00a0 \u00a0 and\u00a0 \u00a0 \\(1\\over y\\) = v, then the equation becomes<\/p>\n

\\(1\\over 2\\)u + \\(1\\over 3\\)v = 2\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 3u + 2v = 12\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n

\\(1\\over 3\\)u + \\(1\\over 2\\)v = \\(13\\over 6\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a02u + 3v = 13\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………….(2)<\/p>\n

Multiply the equation (1) by 3 and equation (2) by 2, we get<\/p>\n

9u + 6v = 36\u00a0 \u00a0 \u00a0 \u00a0……(3)<\/p>\n

4u + 6v = 26\u00a0 \u00a0 \u00a0 \u00a0……..(4)<\/p>\n

Subtract equation (4) from equation ( 3), we get<\/p>\n

5u = 10\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 u = 2<\/p>\n

Putting the value of u in equation (3), we get<\/p>\n

18 + 6v = 36\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = 3<\/p>\n

Now,\u00a0 \\(1\\over x\\) = u\u00a0 \\(\\implies\\)\u00a0 x = \\(1\\over 2\\)<\/strong><\/p>\n

and \\(1\\over y\\) = v\u00a0 \\(\\implies\\)\u00a0 y = \\(1\\over 3\\)<\/strong><\/p>\n

(ii)<\/strong>\u00a0 Put \\(1\\over \\sqrt{x}\\) = u\u00a0 \u00a0 and\u00a0 \u00a0 \\(1\\over \\sqrt{y}\\) = v, then the equation becomes<\/p>\n

2u + 3v = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……(1)<\/p>\n

and\u00a0 \u00a0 4u – 9u = -1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……(2)<\/p>\n

Multiply equation (1) by 3 and adding it with equation (2), we get<\/p>\n

10u = 5\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0u = \\(1\\over 2\\)<\/p>\n

Put the value of u in equation (1), we get<\/p>\n

3v = 1\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = \\(1\\over 3\\)<\/p>\n

Now,\u00a0 \\(1\\over \\sqrt{x}\\) = u\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 x = 4<\/strong><\/p>\n

and \\(1\\over \\sqrt{y}\\) = v\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 9<\/strong><\/p>\n

(iii)<\/strong>\u00a0 The given linear equations are<\/p>\n

\\(4\\over x\\) + 3y = 14\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(1)<\/p>\n

and\u00a0 \u00a0 \u00a0\\(3\\over x\\) – 4y = 23\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n

Multiply equation (1) by 4 and equation (2) by 3, we get<\/p>\n

\\(16\\over x\\) + 12y = 56\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(3)<\/p>\n

\\(9\\over x\\) – 12y = 69\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(4)<\/p>\n

Now, Add equation (3) and equation (3), we get<\/p>\n

\\(25\\over x\\) = 125\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = \\(1\\over 5\\)<\/p>\n

Put the value x in equation (1), we get<\/p>\n

3y = -6\u00a0 \\(\\implies\\)\u00a0 y = -2<\/p>\n

Hence, x = \\(1\\over 5\\)\u00a0 and\u00a0 y = -2.<\/strong><\/p>\n

(iv)<\/strong>\u00a0 Let u = \\(1\\over x – 1\\)\u00a0 and\u00a0 v = \\(1\\over y – 2\\). Then, the given linear equations becomes,<\/p>\n

5u + v = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n

and\u00a0 \u00a0 6u – 3v = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n

Multiply equation (1) by 3 and adding it to equation (2), we get<\/p>\n

21 u = 7\u00a0 \u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 u = \\(1\\over 3\\)<\/p>\n

Put the value of u in equation (1), we get<\/p>\n

\\(5\\over 3\\) + v = 2\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = \\(1\\over 3\\)<\/p>\n

Now, u = \\(1\\over x – 1\\)\u00a0 \\(\\implies\\)\u00a0 x = 4<\/p>\n

and v = \\(1\\over y – 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 y = 5<\/p>\n

Hence, x = 4 and y = 5.<\/strong><\/p>\n

(v)\u00a0<\/strong> The given linear equations are :<\/p>\n

\\(7x – 2y\\over xy\\)\u00a0 = 5\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0\\(7\\over y\\) – \\(2\\over x\\) = 5<\/p>\n

and\u00a0 \u00a0 \u00a0\\(8x + 7y\\over xy\\)\u00a0 =\u00a0 15\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0\\(8\\over y\\) + \\(7\\over x\\) = 15<\/p>\n

Let u = \\(1\\over x\\)\u00a0 and\u00a0 v = \\(1\\over y\\). Then, the above equations becomes<\/p>\n

7v – 2u\u00a0 = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(1)<\/p>\n

and\u00a0 8v + 7u = 15\u00a0 \u00a0 \u00a0 \u00a0 ………(2)<\/p>\n

Multiply equation (1) by 7 and equation (2) by 2, we get<\/p>\n

49 – 14u = 35\u00a0 \u00a0 \u00a0 \u00a0 ……..(3)<\/p>\n

and\u00a0 16v + 14u = 30\u00a0 \u00a0 \u00a0 \u00a0 ……..(4)<\/p>\n

Add equation (3) and equation (4), we get<\/p>\n

65v = 65\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0v = 1<\/p>\n

Put the value of v in equation (1), we get<\/p>\n

7 – 2u = 5\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 u = 1<\/p>\n

Now, u = \\(1\\over x\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 x = 1<\/p>\n

and v = \\(1\\over y\\)\u00a0 \u00a0 \\(\\implies\\)\u00a0 y = 1<\/p>\n

Hence, x = 1, y = 1<\/strong>.<\/p>\n

(vi)<\/strong>\u00a0 The given linear equations are<\/p>\n

6x + 3y = 6xy<\/p>\n

and\u00a0 \u00a0 \u00a0 \u00a02x + 4y = 5xy<\/p>\n

On dividing above equations by xy, we get<\/p>\n

\\(3\\over x\\)\u00a0 +\u00a0 \\(6\\over y\\)\u00a0 = 6<\/p>\n

and\u00a0 \u00a0\\(4\\over x\\) + \\(2\\over y\\) = 5<\/p>\n

Put \\(1\\over x\\) = u and \\(1\\over y\\) = u, the above equation becomes,<\/p>\n

3u + 6v = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n

and\u00a0 4u + 2v = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n

Multiply equation (2) by 3 and subtracting it from equation (1), we get<\/p>\n

-9u = -9\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0u = 1<\/p>\n

Put the value of u in equation (1), we get<\/p>\n

6v = 3\u00a0 \\(\\implies\\)\u00a0 \u00a0v = \\(1\\over 2\\)<\/p>\n

Now,\u00a0 \\(1\\over x\\) = u\u00a0 \\(\\implies\\)\u00a0 x = 1<\/p>\n

\\(1\\over y\\) = v\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 2<\/p>\n

Hence,\u00a0 x = 1 and y = 2<\/strong>.<\/p>\n

(vii)<\/strong>\u00a0 The given linear equations are<\/p>\n

\\(10\\over x + y\\) + \\(2\\over x – y\\) = 4<\/p>\n

and\u00a0 \u00a0 \u00a0\\(15\\over x + y\\) – \\(5\\over x – y\\) = -2<\/p>\n

Put u = \\(1\\over x + y\\)\u00a0 and v = \\(1\\over x – y\\), the given equations becomes<\/p>\n

10u + 2v = 4\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a05u + v = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n

and 15u – 5v = -2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………(2)<\/p>\n

Multiply equation (1) by 5 and adding it with equation (2), we get<\/p>\n

40u = 8 \\(\\implies\\) u = \\(1\\over 5\\)<\/p>\n

Put the value of u in equation (1), we get<\/p>\n

v = 1<\/p>\n

Now, u = \\(1\\over x + y\\)\u00a0 \\(\\implies\\)\u00a0 \u00a0x + y = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(4)<\/p>\n

and v = \\(1\\over x – y\\)\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0x – y = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(5)<\/p>\n

Adding equation (4) and equation (5), we get<\/p>\n

2x = 6 \\(\\implies\\)\u00a0 x = 3<\/p>\n

Put the value of x in equation (4), we get<\/p>\n

3 + y = 5\u00a0 \u00a0 \\(\\implies\\)\u00a0 y = 2<\/p>\n

Hence, x = 3 and y = 2<\/strong>.<\/p>\n

(viii)<\/strong>\u00a0 The given equation are<\/p>\n

\\(1\\over 3x + y\\) + \\(1\\over 3x – y\\) = \\(3\\over 4\\)<\/p>\n

and\u00a0 \u00a0\\(1\\over 2(3x + y)\\) – \\(1\\over 2(3x – y)\\) = \\(-1\\over 8\\)<\/p>\n

Putting u = \\(1\\over 3x + y\\) and v = \\(1\\over 3x – y\\), then the equation becomes,<\/p>\n

u + v = \\(3\\over 4\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n

and \\(1\\over 2\\)u – \\(1\\over 2\\)v = \\(-1\\over 8\\)<\/p>\n

\\(\\implies\\)\u00a0 u – v = \\(-1\\over 4\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(2)<\/p>\n

Add equation (1) and equation (2), we get<\/p>\n

2u = \\(1\\over 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0u = \\(1\\over 4\\)<\/p>\n

Put the value of u in equation (1), we get<\/p>\n

\\(1\\over 4\\) + v = \\(3\\over 4\\)\u00a0 \\(\\implies\\)\u00a0 v = \\(1\\over 2\\)<\/p>\n

Now, u = \\(1\\over 3x + y\\)\u00a0 \\(\\implies\\)\u00a0 \u00a03x + y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(3)<\/p>\n

and\u00a0 v = \\(1\\over 3x – y\\)\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x – y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(4)<\/p>\n

Adding equation (3) and equation (4), we get<\/p>\n

6x = 6\u00a0 \u00a0or\u00a0 x = 1<\/p>\n

Put the value x in equation (3), we get<\/p>\n

3 + y = 4\u00a0 \u00a0 or\u00a0 \u00a0 \u00a0y = 1<\/p>\n

Hence, x = 1, y = 1<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Solve the following pairs of equations by reducing them to a pair of linear of linear equations : (i)\u00a0 \\(1\\over 2x\\) + \\(1\\over 3y\\) = 2\u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \\(1\\over 3x\\) + \\(1\\over 2y\\) = 2 (ii)\u00a0 \\(2\\over \\sqrt{x}\\) + \\(3\\over \\sqrt{y}\\) = 2\u00a0 \u00a0 \u00a0 \u00a0and\u00a0 \u00a0 \\(4\\over \\sqrt{x}\\) – \\(9\\over \\sqrt{y}\\) …<\/p>\n

Solve the following pairs of equations by reducing them to a pair of linear of linear equations<\/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 pairs of equations by reducing them to a pair of linear of linear equations - 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-pairs-of-equations-by-reducing-them-to-a-pair-of-linear-of-linear-equations\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solve the following pairs of equations by reducing them to a pair of linear of linear equations - 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