{"id":11094,"date":"2022-06-18T04:35:58","date_gmt":"2022-06-17T23:05:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11094"},"modified":"2022-06-18T04:36:02","modified_gmt":"2022-06-17T23:06:02","slug":"solve-the-following-pair-of-linear-equations-by-the-elimination-method-and-the-substitution-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-the-following-pair-of-linear-equations-by-the-elimination-method-and-the-substitution-method\/","title":{"rendered":"Solve the following pair of linear equations by the elimination method and the substitution method"},"content":{"rendered":"

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

(i)<\/strong>\u00a0 x + y = 5\u00a0 and\u00a0 \u00a02x – 3y = 4<\/p>\n

(ii)<\/strong>\u00a0 3x + 4y = 10\u00a0 and\u00a0 2x – 2y = 2<\/p>\n

(iii)<\/strong>\u00a0 3x – 5y – 4 = 0\u00a0 and\u00a0 9x = 2y + 7<\/p>\n

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

Solution :<\/h2>\n

(i)\u00a0 By Elimination Method :<\/strong><\/p>\n

The given equations are<\/p>\n

x + y = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 \u00a02x – 3y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(2)<\/p>\n

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

5x = 19\u00a0 \u00a0\\(\\implies\\)\u00a0 x = \\(19\\over 5\\)<\/p>\n

Now, Put the\u00a0 value of x in equation (1), we get<\/p>\n

\\(19\\over 5\\) + y = 5\u00a0 \u00a0 \\(\\implies\\)\u00a0 y = \\(6\\over 5\\)<\/p>\n

Hence, x = \\(19\\over 5\\) and y = \\(6\\over 5\\)<\/strong><\/p>\n

By Substitution Method :\u00a0<\/strong><\/p>\n

The given equations are<\/p>\n

x + y = 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 \u00a02x – 3y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(2)<\/p>\n

From equation (1),\u00a0 y = 5 – x<\/p>\n

Substituting the value of y in equation (2), we get<\/p>\n

2x – 15 + 3x = 4\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a05x = 4 + 19<\/p>\n

\\(\\implies\\)\u00a0 5x = 19\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x = \\(19\\over 5\\)<\/p>\n

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

\\(19\\over 5\\) + y = 5\u00a0 \u00a0 \\(\\implies\\)\u00a0 y = \\(6\\over 5\\)<\/p>\n

Hence, x = \\(19\\over 5\\) and y = \\(6\\over 5\\)<\/strong><\/p>\n

(i)\u00a0 By Elimination Method :<\/strong><\/p>\n

The given equations are<\/p>\n

3x + 4y = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 \u00a02x – 2y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n

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

7x = 14\u00a0 \u00a0\\(\\implies\\)\u00a0 x = 2<\/p>\n

Now, Put the\u00a0 value of x in equation (1), we get<\/p>\n

3(2) + 4y = 10\u00a0 \u00a0 \\(\\implies\\)\u00a0 4y = 10 – 6 = 4\u00a0 \u00a0\\(\\implies\\)\u00a0 y = 1<\/p>\n

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

By Substitution Method :\u00a0<\/strong><\/p>\n

The given equations are<\/p>\n

3x + 4y = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 \u00a02x – 2y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(2)<\/p>\n

From equation (2),\u00a0 y = x – 1<\/p>\n

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

3x + 4(x – 1) = 10<\/p>\n

\\(\\implies\\)\u00a0 7x = 14\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x = 2<\/p>\n

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

3(2) + 4y = 10\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 y = 1<\/p>\n

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

(iii)\u00a0 By Elimination Method<\/strong> :<\/p>\n

The given equations are<\/p>\n

3x – 5y – 4 = 0\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x – 5y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …..(1)<\/p>\n

and\u00a0 9x = 2y + 7\u00a0 \\(\\implies\\)\u00a0 9x – 2y = 7\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n

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

-13y = 5\u00a0 \u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 \u00a0 \u00a0 y = \\(-5\\over 13\\)<\/p>\n

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

3x – 5(\\(-5\\over 13\\)) = 4<\/p>\n

\\(\\implies\\)\u00a0 \u00a03x = \\(52 – 25\\over 13\\)<\/p>\n

\\(\\implies\\)\u00a0 3x = \\(27\\over 13\\)<\/p>\n

\\(\\implies\\)\u00a0 x = \\(9\\over 13\\)<\/p>\n

Hence, x = \\(9\\over 13\\)\u00a0 and\u00a0 y = \\(-5\\over 13\\)<\/strong><\/p>\n

By Substitution Method :<\/strong><\/p>\n

The given equations are<\/p>\n

3x – 5y – 4 = 0\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x – 5y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …..(1)<\/p>\n

and\u00a0 9x = 2y + 7\u00a0 \\(\\implies\\)\u00a0 9x – 2y = 7\u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n

From equation (2),\u00a0 \u00a0 y = \\(9x – 7\\over 2\\)<\/p>\n

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

3x – 5(\\(9x – 7\\over 2\\)) = 4\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 6x – 45x + 35 = 8<\/p>\n

\\(\\implies\\)\u00a0 -39x = 8 – 35\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0-39x = -27<\/p>\n

\\(\\implies\\)\u00a0 x = \\(9\\over 13\\)<\/p>\n

Now, Put the value of x in equation (2), we get<\/p>\n

3 \\(\\times\\) \\(9\\over 13\\) – 5y = 4<\/p>\n

\\(\\implies\\)\u00a0 5y = \\(-25\\over 13\\)<\/p>\n

\\(\\implies\\)\u00a0 y = \\(-5\\over 13\\)<\/p>\n

Hence, x = \\(9\\over 13\\)\u00a0 and\u00a0 y = \\(-5\\over 13\\)<\/strong><\/p>\n

(iv)\u00a0 By Elimination Method :<\/strong><\/p>\n

The given equations are<\/p>\n

\\(x\\over 2\\) + \\(2y\\over 3\\) = -1\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x + 4y = -6\u00a0 \u00a0 \u00a0 \u00a0………(i)<\/p>\n

and\u00a0 x – \\(y\\over 3\\) = 3\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 \u00a0 \u00a03x – y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(ii)<\/p>\n

Multiplying equation (2) by\u00a0 4 and adding to equation (1), we get<\/p>\n

15x = 30\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0x = 2<\/p>\n

Now, put\u00a0 x = 2 in equation (2), we get<\/p>\n

3(2) – y\u00a0 = 9\u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0– y = 9 – 6 = 3<\/p>\n

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

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

By Substitution Method :<\/strong><\/p>\n

The given equations are<\/p>\n

\\(x\\over 2\\) + \\(2y\\over 3\\) = -1\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a03x + 4y = -6\u00a0 \u00a0 \u00a0 \u00a0………(i)<\/p>\n

and\u00a0 x – \\(y\\over 3\\) = 3\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 \u00a0 \u00a03x – y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(ii)<\/p>\n

From equation (2),\u00a0 y = 3x – 9<\/p>\n

Putting the value of y in equation (1), we get\u00a0 y = 3x – 9<\/p>\n

3x + 4(3x – 9) = -6\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 3x + 12x – 36 = -6<\/p>\n

\\(\\implies\\)\u00a0 \u00a015x = 30\u00a0 \u00a0\\(\\implies\\)\u00a0 x = 2<\/p>\n

Putting the value of x in equation (2), we get<\/p>\n

3(2) – y = 9\u00a0\u00a0\\(\\implies\\)\u00a0 \u00a0– y = 9 – 6 = 3<\/p>\n

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

Hence,\u00a0x = 2 and y = -3<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Solve the following pair of linear equations by the elimination method and the substitution method : (i)\u00a0 x + y = 5\u00a0 and\u00a0 \u00a02x – 3y = 4 (ii)\u00a0 3x + 4y = 10\u00a0 and\u00a0 2x – 2y = 2 (iii)\u00a0 3x – 5y – 4 = 0\u00a0 and\u00a0 9x = 2y + …<\/p>\n

Solve the following pair of linear equations by the elimination method and 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 elimination method and 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-elimination-method-and-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 elimination method and the substitution method - 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