{"id":1943,"date":"2022-04-23T09:48:03","date_gmt":"2022-04-23T09:48:03","guid":{"rendered":"https:\/\/hbsesolutions.com\/?p=1943"},"modified":"2022-04-23T09:48:43","modified_gmt":"2022-04-23T09:48:43","slug":"hbse-8th-class-maths-solutions-chapter-2-ex-2-1","status":"publish","type":"post","link":"https:\/\/hbsesolutions.com\/hbse-8th-class-maths-solutions-chapter-2-ex-2-1\/","title":{"rendered":"HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1"},"content":{"rendered":"

Haryana State Board HBSE 8th Class Maths Solutions<\/a>\u00a0Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.<\/p>\n

Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.1<\/h2>\n

Question 1.
\nSolve the following equations:
\n1. x – 2 = 7
\n2. y + 3 = 10
\n3. 6 = z + 2
\n4. \\(\\frac{3}{7}\\) + x = \\(\\frac{17}{7}\\)
\n5. 6x = 12
\n7. \\(\\frac{2x}{3}\\) = 18
\n8. 1.6 = \\(\\frac{y}{1.5}\\)
\n9. 7x – 9 = 16
\n10. 14y – 8 = 13
\n11. 17 + 6P = 9
\n12. \\(\\frac{x}{3}\\) + 1 = \\(\\frac{7}{15}\\)
\nSolution:
\n1. x – 2 = 7
\nor x – 2 + 2 = 7 + 2 (adding 2 both sides)
\n\u2234 x = 9<\/p>\n

2. y + 3 = 10
\ny + 3 – 3 = 10 – 3 (subtract 3 from both sides)
\n\u2234 y = 7<\/p>\n

3. 6 = x + 2
\n6 – 2 = z + 2 – 2 (subtract 2 from both sides)
\n4 = z
\n\u2234 z = 4<\/p>\n

4. \\(\\frac{3}{7}\\) + x = \\(\\frac{17}{7}\\)
\nx = \\(\\frac{17}{7}\\) – \\(\\frac{3}{7}\\) = \\(\\frac{14}{7}\\) = 2
\n\u2234 x = 2<\/p>\n

5. 6x = 12
\nor \\(\\frac{6x}{6}\\) = \\(\\frac{12}{6}\\) (divide both side by 6)
\n\u2234 x = 2<\/p>\n

\"HBSE<\/p>\n

6. \\(\\frac{t}{5}\\) = 10
\nor \\(\\frac{t}{5}\\) \u00d7 5 = 10 \u00d7 5 (multiply both sides by 5)
\n\u2234 t = 50<\/p>\n

7. \\(\\frac{2x}{3}\\) = 18
\nor 2x = 18 \u00d7 3
\n\u2234 x = \\(\\frac{18 \\times 3}{2}\\) = 9 \u00d7 3 = 27
\n\u2234 x = 27<\/p>\n

8. 1.6 = \\(\\frac{y}{1.5}\\)
\nor y = 1.6 \u00d7 1.5 = 2.40
\n\u2234 y = 2.4<\/p>\n

9. 7x – 9 = 16
\nor 7x – 9 + 9 = 16 + 9 (adding 9 both sides)
\nor 7x = 25
\n\u2234 x = \\(\\frac{25}{7}\\)<\/p>\n

\"HBSE<\/p>\n

10. 14y – 8 = 13
\n14y – 8 + 8 = 13 + 8 (adding 8 both sides)
\nor 14y = 21
\nor y = \\(\\frac{21}{14}\\) = \\(\\frac{3}{2}\\)
\n\u2234 y = \\(\\frac{3}{2}\\)<\/p>\n

11. 17 + 6P = 9
\n17 + 6P – 17 = 9 – 17 (subtract -17 from both sides)
\nor 6P = -8
\nor P = \\(\\frac{-8}{6}\\) = \\(\\frac{-4}{3}\\)
\n\u2234 P = \\(\\frac{-4}{3}\\)<\/p>\n

12. \\(\\frac{x}{3}\\) + 1 = \\(\\frac{7}{15}\\)
\nor \\(\\frac{x}{3}\\) + 1 – 1 = \\(\\frac{7}{15}\\) – 1 (subtract 1 from both sides)
\nor \\(\\frac{x}{3}\\) = \\(\\frac{7-15}{15}\\)
\nor \\(\\frac{x}{3}\\) = \\(\\frac{-8}{15}\\)
\n\u2234 x = \\(\\frac{-8}{15}\\) \u00d7 3 = \\(\\frac{-8}{5}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Haryana State Board HBSE 8th Class Maths Solutions\u00a0Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers. Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.1 Question 1. Solve the following equations: 1. x – 2 = 7 2. y + 3 = 10 3. …<\/p>\n

HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"","spay_email":"","footnotes":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/posts\/1943"}],"collection":[{"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/comments?post=1943"}],"version-history":[{"count":1,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/posts\/1943\/revisions"}],"predecessor-version":[{"id":1944,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/posts\/1943\/revisions\/1944"}],"wp:attachment":[{"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/media?parent=1943"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/categories?post=1943"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/hbsesolutions.com\/wp-json\/wp\/v2\/tags?post=1943"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}