{"id":220,"date":"2022-04-13T10:21:45","date_gmt":"2022-04-13T10:21:45","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=220"},"modified":"2025-01-31T06:45:23","modified_gmt":"2025-01-31T06:45:23","slug":"k-6-solve-rational-equations","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/k-6-solve-rational-equations\/","title":{"rendered":"K.6 Solve rational equations"},"content":{"rendered":"\n

Solve rational equations<\/strong><\/h2>\n\n\n\n

Key Notes :<\/p>\n\n\n\n

To solve a rational equation, first clear the fractions, by multiplying both sides by the denominators or by the lowest common denominator (LCD). Then solve for the variable.<\/p>\n\n\n\n

Learn with an example<\/strong><\/p>\n\n\n\n

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Solve for w<\/em>.<\/strong><\/p>\n\n\n\n

-4\/w-8 = -2\/w+1<\/p>\n\n\n\n

There may be 1 or 2 solutions.<\/strong><\/p>\n\n\n\n

w = ________ or w = ___________<\/p>\n<\/div><\/div>\n\n\n\n

Solve for w<\/em>.<\/p>\n\n\n\n

-4\/w-8 = -2\/w+1<\/p>\n\n\n\n

-4[(w-8)(w+1)]\/w-8 = -2[(w-8)(w+1)]\/w+1<\/em> Multiply both sides by (w \u2212 8)(w + 1)<\/p>\n\n\n\n

-4(w+1) = -2(w-8) Simplify<\/p>\n\n\n\n

-4w-4 = -2w+16 Apply the distributive property<\/p>\n\n\n\n

-2w-4 = 16 Add 2w to both sides<\/p>\n\n\n\n

-2w = 20 Add 4 to both sides<\/p>\n\n\n\n

w = -10 Divide both sides by \u2013<\/sup>2<\/p>\n\n\n\n

Now check whether this is an extraneous solution. Plugging w = -10 into the first denominator, w-8 , yields -18 . Plugging w = -10  into the second denominator, w+1 , yields -9 .Since neither denominator is 0, which would be undefined, this is a valid solution.<\/p>\n\n\n\n

The solution is w = -10<\/p>\n<\/div><\/div>\n\n\n\n

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Solve for u.<\/strong><\/p>\n\n\n\n

8\/u+3 = 3\/u-2<\/p>\n\n\n\n

There may be 1 or 2 solutions.<\/strong><\/p>\n\n\n\n

u = ________ or u = ___________<\/p>\n<\/div><\/div>\n\n\n\n

Solve for u<\/em>.<\/p>\n\n\n\n

8\/u+3 = 3\/u-2<\/p>\n\n\n\n

8[(u+3)(u-2)]\/u+3 = 3[(u+3)(u-2)]\/u-2 Multiply both sides by (u + 3)(u \u2212 2)<\/p>\n\n\n\n

8(u-2) = 3(u+3) Simplify<\/p>\n\n\n\n

8u-16 = 3u+9 Apply the distributive property<\/p>\n\n\n\n

5u-16 = 9 Subtract 3u from both sides<\/p>\n\n\n\n

5u = 25 Add 16 to both sides<\/p>\n\n\n\n

u = 5 Divide both sides by 5<\/p>\n\n\n\n

Now check whether this is an extraneous solution. Plugging u = 5 into the first denominator, u+3 , yields 8. Plugging u = 5 into the second denominator, u-2, yields 3  Since neither denominator is 0, which would be undefined, this is a valid solution.<\/p>\n\n\n\n

The solution is u = 5 <\/p>\n<\/div><\/div>\n\n\n\n

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Solve for k<\/em>.<\/strong><\/p>\n\n\n\n

k-8\/6 = k-10\/7<\/p>\n\n\n\n

There may be 1 or 2 solutions.<\/strong><\/p>\n\n\n\n

k = ________ or k= ___________<\/p>\n<\/div><\/div>\n\n\n\n

Solve for k<\/em>.<\/p>\n\n\n\n

k-8\/6 = k-10\/7<\/p>\n\n\n\n

(k-8)(6 . 7) \/ 6 = (k-10)(6 .7)\/7 Multiply both sides by 6 \u00b7 7<\/p>\n\n\n\n

7(k-8) = 6(k-10) Simplify<\/p>\n\n\n\n

7k-56 = 6k-60 Apply the distributive property<\/p>\n\n\n\n

k-56 = -60 Subtract 6k from both sides<\/p>\n\n\n\n

k = -4 Add 56 to both sides<\/p>\n\n\n\n

Now check whether this is an extraneous solution. Since neither denominator is 0, which would be undefined, this is a valid solution.<\/p>\n\n\n\n

The solution is k = -4<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!<\/p>\n\n\n\n

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\"\"<\/a><\/figure>\n<\/div>\n\n\n\n
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\"\"<\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Solve rational equations Key Notes : To solve a rational equation, first clear the fractions, by multiplying both sides by the denominators or by the lowest common denominator (LCD). Then solve for the variable. Learn with an example Solve for w. -4\/w-8 = -2\/w+1 There may be 1 or 2 solutions. w = ________ or wContinue reading “K.6 Solve rational equations”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-220","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/220","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=220"}],"version-history":[{"count":16,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/220\/revisions"}],"predecessor-version":[{"id":17487,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/220\/revisions\/17487"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=220"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}